Concept of Physics | H C Verma | Chapter 8 : Work and Energy | Solution

1. A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown (a) vertically downward (b) vertically upward (c) horizontally (d) the speed does not depend on the initial direction.

Answer: (d) the speed does not depend on the initial direction.

Solution:

1. Initial Energy: The stone initially has both kinetic energy
   (K_i = ½mv²) and gravitational potential energy
   (U_i = mgh) relative to the ground.
2. Final Energy: When the stone hits the ground, its potential energy is zero
   (U_f = 0), and all its initial energy is converted into kinetic energy
   (K_f = ½mv_f²).
3. Conservation Equation:
   ½mv² + mgh = ½mv_f²
4. Conclusion: The final speed v_f depends only on the initial speed v and the height h. Since energy is a scalar quantity, the direction of projection does not affect the final speed.

2. Two springs A and B (k_A = 2k_B) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is (a) E/2 (b) 2E (c) E (d) E/4.

Answer: (b) 2E

Solution:

1. Energy Formula: The energy stored in a spring is
   U = ½kx².
2. Using Force Relation: Since F = kx, we get
   x = F/k.
3. Substitute:
   U = ½k(F/k)² = F²/(2k)
4. Comparison: For equal force,
   • E_A = F²/(2k_A)
   • E_B = F²/(2k_B)
5. Result: Since k_A = 2k_B,E_B/E_A = k_A/k_B = 2⇒ E_B = 2E

3 Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is
(a) ½kx² (b) −½kx² (c) ¼kx² (d) −¼kx²

Answer: (d) −¼kx²

Solution: When a spring is stretched by a total length x, the total work done by the spring is −½kx². Because the two masses are equal and are pulled symmetrically, the work done by the spring is divided equally between the two, resulting in −¼kx² of work being done on each mass.

4 The negative of the work done by the conservative internal forces on a system equals the change in
(a) total energy (b) kinetic energy (c) potential energy (d) none of these.

Answer: (c) potential energy

Solution: The change in potential energy (ΔU) of a system is defined as the negative of the work done by the conservative internal forces acting within that system.

5 The work done by the external forces on a system equals the change in
(a) total energy (b) kinetic energy (c) potential energy (d) none of these.

Answer: (a) total energy

Solution: The work done by external forces acting on a system results in a change in the total mechanical energy (sum of kinetic and potential energy) of that system.

6 The work done by all the forces (external and internal) on a system equals the change in
(a) total energy (b) kinetic energy (c) potential energy (d) none of these.

Answer: (b) kinetic energy

Solution: According to the work-energy theorem, the total work done by all forces acting on a system equals the change in the total kinetic energy of the system.

7 ————— of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is
(a) Kinetic energy (b) Total mechanical energy (c) Potential energy (d) Total energy.

Answer: (c) Potential energy

Solution: Potential energy is a property determined by the configuration of a system. For a system consisting of two particles, the potential energy depends only on the separation between them.

8 A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be
(a) zero (b) mgvt cos²θ (c) mgvt sin²θ (d) mgvt sin 2θ.

Answer: (c) mgvt sin²θ

Solution: Since the block does not slide on the incline and the elevator moves with a uniform velocity v, the block is in equilibrium. The component of gravity down the incline is balanced by friction, so f = mg sinθ. The displacement of the block in time t is vertical, s = vt. The angle between friction (along the incline) and displacement (vertical) is (90° − θ). Hence, work done by friction is W = f · s · cos(90° − θ) = (mg sinθ)(vt)(sinθ) = mgvt sin²θ.

9 A block of mass m slides down a smooth vertical circular track. During the motion, the block is in
(a) vertical equilibrium (b) horizontal equilibrium (c) radial equilibrium (d) none of these.

Answer: (d) none of these

Solution: As the block moves along the circular track, its velocity changes in both magnitude and direction. Hence, it has both tangential and radial acceleration. Since acceleration is present, the net force is not zero in any direction, so the block is not in equilibrium.

10 A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is
(a) √gl (b) √2gl (c) √3gl (d) √5gl.

Answer: (c) √3gl

Solution: For the particle to complete the vertical circle, the minimum speed at the top must be √(gl). Using conservation of mechanical energy between the horizontal position and the top:
½mv² = ½m(gl) + mgl = 3/2 mgl.
Thus, v² = 3gl and v = √(3gl).

OBJECTIVE II

1 A heavy stone is thrown from a cliff of height h in a given direction. The speed with which it hits the ground
(a) must depend on the speed of projection (b) must be larger than the speed of projection (c) must be independent of the speed of projection (d) may be smaller than the speed of projection.

Answer: (a), (b)

Solution: Using conservation of mechanical energy, ½mv_p² + mgh = ½mv_f², which gives v_f = √(v_p² + 2gh). Hence the final speed depends on the speed of projection and is greater than it.

2 The total work done on a particle is equal to the change in its kinetic energy
(a) always (b) only if the forces acting on it are conservative (c) only if gravitational force alone acts on it (d) only if elastic force alone acts on it.

Answer: (a) always

Solution: This is the work-energy theorem which holds for all types of forces, whether conservative or non-conservative.

3 A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that
(a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a circular path.

Answer: (c), (d)

Solution: Since the force is perpendicular to velocity, work done is zero, so kinetic energy remains constant. A constant perpendicular force results in uniform circular motion.

4 Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance l on a rough surface. The following quantities will be same as observed by the two observers
(a) kinetic energy of the block at time t (b) work done by friction (c) total work done on the block (d) acceleration of the block.

Answer: (d) acceleration of the block

Solution: Acceleration is invariant in all inertial frames moving with constant relative velocity, while quantities like work and kinetic energy depend on the frame.

5 You lift a suitcase from the floor and keep it on a table. The work done by you on the suitcase does not depend on
(a) the path taken by the suitcase (b) the time taken by you in doing so (c) the weight of the suitcase (d) your weight.

Answer: (a), (b), (d)

Solution: Work done against gravity depends only on vertical displacement (not path), is independent of time, and depends on the suitcase’s weight (mg), not on the person’s weight.

6 No work is done by a force on an object if
(a) the force is always perpendicular to its velocity (b) the force is always perpendicular to its acceleration (c) the object is stationary but the point of application of the force moves on the object (d) the object moves in such a way that the point of application of the force remains fixed.

Answer: (a), (c), (d)

Solution: Work is given by the dot product of force and displacement. If the force is perpendicular to velocity (and hence displacement), the work is zero. Also, no work is done if the object remains stationary even if the point of application moves, or if the point of application remains fixed while the object moves.

7 A particle of mass m is attached to a light string of length l, the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle.
(a) The string becomes slack when the particle reaches its highest point. (b) The velocity of the particle becomes zero at the highest point. (c) The kinetic energy of the ball in initial position was ½mv² = mgl. (d) The particle again passes through the initial position.

Answer: (a), (d)

Solution: To just complete the circle, the tension becomes zero at the highest point, so the string becomes slack. The velocity at the highest point is not zero; it must be √(gl). Since the motion is circular, the particle again passes through the initial position.

8 The kinetic energy of a particle continuously increases with time.
(a) The resultant force on the particle must be parallel to the velocity at all instants. (b) The resultant force on the particle must be at an angle less than 90° all the time. (c) Its height above the ground level must continuously decrease. (d) The magnitude of its linear momentum is increasing continuously.

Answer: (b), (d)

Solution: For kinetic energy to increase, the work done must be positive, which requires the force to have a component along the velocity, so the angle must be less than 90°. Since kinetic energy increases, the magnitude of momentum also increases.

9 One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is ½kx². The possible cases are
(a) the spring was initially compressed by a distance x and was finally in its natural length (b) it was initially stretched by a distance x and finally was in its natural length (c) it was initially in its natural length and finally in a compressed position (d) it was initially in its natural length and finally in a stretched position.

Answer: (a), (b)

Solution: Work done by the spring is W = ½k(x_i² − x_f²). To get positive work ½kx², the spring must move from displacement x to natural length, so initial displacement is x and final is zero.

10 A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1 s.
(a) The tension in the string is Mg. (b) The tension in the string is F. (c) The work done by the tension on the block is 20 J in the above 1 s. (d) The work done by the force of gravity is –20 J in the above 1 s.

Answer: (b)

Solution: For a light string and smooth pulley, the tension is the same throughout the string, so T = F. The increase in kinetic energy is due to net work, not just the work done by tension alone.

EXERCISES

1. The mass of cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6.0 km/h to 12 km/h.

Solution:
Mass (m) = 90 kg
Initial speed (v₁) = 6.0 km/h = 5/3 m/s
Final speed (v₂) = 12 km/h = 10/3 m/s
Increase in kinetic energy:
ΔK = (1/2) m (v₂² − v₁²)
ΔK = (1/2 × 90) [(10/3)² − (5/3)²]
ΔK = 45 × (100/9 − 25/9) = 45 × (75/9) = 375 J

2. A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s² for 5.00 s. Compute its final kinetic energy.

Solution:
Initial velocity (u) = 10.0 m/s
Acceleration (a) = 3.00 m/s²
Time (t) = 5.00 s
Final velocity:
v = u + at = 10 + (3 × 5) = 25 m/s
Final kinetic energy:
K = (1/2) m v² = (1/2 × 2 × 25²) = 625 J

3. A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force?

Solution:
Force = 100 N (opposite direction)
Displacement = 4.0 m
Work done:
W = Fd cos 180° = 100 × 4 × (−1) = −400 J

4. A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

Solution:
Mass (m) = 5.0 kg
Length = 10 m
Angle = 30°
Height h = 10 sin30° = 5 m
Work done by gravity:
W = mgh = 5 × 9.8 × 5 = 245 J

5. A constant force of 2.50 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work done and the average power delivered.

Solution:
Force = 2.50 N
Mass = 0.015 kg
Displacement = 2.50 m
Work done:
W = F × s = 2.50 × 2.50 = 6.25 J

Acceleration:
a = F/m = 2.50 / 0.015 = 500/3 m/s²
Using s = (1/2)at²:
2.50 = (1/2 × 500/3) t²
t² = 0.03 → t = √0.03 s

Average power:
P = W/t = 6.25 / √0.03 ≈ 36.1 W

6. A particle moves from a point r₁ = (2 m)i + (3 m)j to another point r₂ = (3 m)i + (2 m)j during which a certain force F = (5 N)i + (5 N)j acts on it. Find the work done by the force on the particle during the displacement.

Solution:
Displacement d = r₂ − r₁ = (3−2)i + (2−3)j = (1)i − (1)j
Force F = 5i + 5j
Work done:
W = F · d = (5i + 5j) · (1i − 1j)
W = 5(1) + 5(−1) = 0 J

7. A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of 0.5 m/s², find the work done by the man on the block during the motion.

Solution:
Mass = 2 kg
Acceleration = 0.5 m/s²
Distance = 40 m
Force applied:
F = ma = 2 × 0.5 = 1 N
Work done:
W = F × d = 1 × 40 = 40 J

8. A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.

Solution:
Work done:
W = ∫₀ᵈ (a + bx) dx
W = [ax + (1/2)bx²]₀ᵈ
W = ad + (1/2)bd²

9. A block of mass 250 g slides down an incline of inclination 37° with a uniform speed. Find the work done against the friction as the block slides through 1.0 m.

Solution:
Mass = 0.25 kg
Distance = 1.0 m
sin37° ≈ 0.6
Friction force:
f = mg sinθ = 0.25 × 10 × 0.6 = 1.5 N
Work done against friction:
W = f × d = 1.5 × 1.0 = 1.5 J

10. A block of mass m is kept over another block of mass M and the system rests on a horizontal surface. A constant horizontal force F acting on the lower block produces an acceleration F / [2(m + M)] in the system, the two blocks always move together.

Solution:
Acceleration a = F / [2(m + M)]

(a) Coefficient of friction:
F − μ(m + M)g = (m + M)a = F/2
μ(m + M)g = F/2
μ = F / [2(m + M)g]

(b) Friction on smaller block:
f = m × a = mF / [2(m + M)]

(c) Work done by friction on smaller block:
W = f × d = mFd / [2(m + M)]

11. A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force.

Solution:
(a) For uniform motion, net force is zero.
Horizontal: F cosθ = μN
Vertical: N = W − F sinθ
Substitute:
F cosθ = μ(W − F sinθ)
F = μW / (cosθ + μ sinθ)
Work done:
W = F cosθ × d = [μW / (cosθ + μ sinθ)] × cosθ × d
W = μWd / (1 + μ tanθ)

Putting values:
W = (0.2 × 2000 × 20) / (1 + 0.2 tanθ)
W = 40000 / (5 + tanθ) J

(b)
Force is minimum when tanθ = μ = 0.2
W = 40000 / (5 + 0.2) = 40000 / 5.2 ≈ 7690 J

12. A block of weight 100 N is slowly slid up on a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.

Solution:
Height gained:
h = d sinθ = 2 × 0.6 = 1.2 m
Work done = increase in potential energy:
W = 100 × 1.2 = 120 J
Same for both (a) and (b)

13. Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h.

Solution:
Initial speed = 20 m/s
Final speed = 0
Using work-energy theorem:
−f × 25 = (1/2 × 500 × (0 − 20²))
−f × 25 = −100000
f = 4000 N

14. Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.

Solution:
Final speed = 20 m/s
F × 25 = (1/2 × 500 × 20²)
F × 25 = 100000
F = 4000 N

15. A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to v = a√x. Find the total work done by all the forces during a displacement from x = 0 to x = d.

Solution:
Initial velocity = 0
Final velocity = a√d
Work done = change in kinetic energy:
W = (1/2) m (a√d)²
W = (1/2) ma²d

16. A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take \(g = 10 \text{ m/s}^2\).

Solution:

(a) The maximum possible acceleration (\(a_{\text{max}}\)) occurs when there is no friction:
\[
a_{\text{max}} = \frac{F – mg \sin \theta}{m} = \frac{20 – 2 \times 10 \times \sin 37^\circ}{2} = \frac{20 – 12}{2} = 4 \text{ m/s}^2
\]
\[
s_{\text{max}} = \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \times 1^2 = 2 \text{ m}
\]
\[
W = F \cdot s = 20 \times 2 = 40 \text{ J}
\]
Thus, the work done by the applied force does not exceed \(40 \text{ J}\).

(b) Work done by gravity:
\[
W_g = -mg \sin \theta \cdot s = -(2 \times 10 \times 0.6) \times 2 = -24 \text{ J}
\]

(c) By the work-energy theorem:
\[
K_f = W_{\text{applied}} + W_g = 40 – 24 = 16 \text{ J}
\]

17. A block of mass 2.0 kg is pushed down an inclined plane of inclination 37° with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s². If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take \(g = 10 \text{ m/s}^2\).

Solution:

\[
s = \frac{1}{2} a t^2 = \frac{1}{2} \times 10 \times 1^2 = 5 \text{ m}
\]

(a)
\[
W_F = F \cdot s = 20 \times 5 = 100 \text{ J}
\]

(b)
\[
W_g = mg \sin \theta \cdot s = (2 \times 10 \times 0.6) \times 5 = 60 \text{ J}
\]

(c)
\[
\Delta K = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 \text{ J}
\]
\[
W_f = 100 – (100 + 60) = -60 \text{ J}
\]

Question 18. A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0.1, how far does the block move before coming to rest ?

Solution:

\[
W_f = \Delta K = 0 – \frac{1}{2} \times 0.25 \times (0.4)^2 = -0.02 \text{ J}
\]
\[
-0.02 = -\mu mg d \Rightarrow d = \frac{0.02}{0.245} \approx 0.0816 \text{ m} \approx 8.2 \text{ cm}
\]

19. Water falling from a 50 m high fall is to be used for generating electric energy. If 1.8 × 10⁵ kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit ?

Solution:

\[
\frac{dm}{dt} = \frac{1.8 \times 10^5}{3600} = 50 \text{ kg/s}
\]
\[
P = \frac{dm}{dt} g h = 50 \times 9.8 \times 50 = 24500 \text{ W}
\]
\[
P_{\text{elec}} = 0.5 \times 24500 = 12250 \text{ W}
\]
\[
N = \frac{12250}{100} \approx 122
\]

20. A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6.0 kg and was at a height of 2.0 m at the time it slipped, how much gravitational potential energy is lost together with the paint ?

Solution:

\[
\Delta U = mgh = 6.0 \times 9.8 \times 2.0 = 117.6 \text{ J} \approx 118 \text{ J}
\]

21. A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.

Solution:

\[
\frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2
\]
\[
v_f = \sqrt{v_i^2 + 2gh} = \sqrt{50^2 + 2 \times 9.8 \times 40} \approx 58 \text{ m/s}
\]

22. The 200 m free style women’s swimming gold medal at Seol Olympic 1988 went to Heike Friendrich… she set a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.

Solution:

\[
v = \frac{200}{117.56} \approx 1.701 \text{ m/s}
\]
\[
F = \frac{P}{v} = \frac{460}{1.701} \approx 270 \text{ N}
\]

23. The US athlete Florence Griffith-Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of 10.54 s. Assume that she achieved her maximum speed in a very short-time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith-Joyner at her full speed. (b) Assuming that the track, the wind etc. offered an average resistance of one tenth of her weight, calculate the work done by the resistance during the run. (c) What power Griffith-Joyner had to exert to maintain uniform speed ?

Solution:

\[
v = \frac{100}{10.54} \approx 9.488 \text{ m/s}
\]
(a)
\[
K = \frac{1}{2} \times 50 \times (9.488)^2 \approx 2250 \text{ J}
\]
(b)
\[
F_r = \frac{1}{10} mg = 49 \text{ N}, \quad W = -49 \times 100 = -4900 \text{ J}
\]
(c)
\[
P = F_r \times v = 49 \times 9.488 \approx 465 \text{ W}
\]

24. A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower the engine should have to do this.

Solution:

\[
P = \frac{dm}{dt} g h = \frac{30}{60} \times 9.8 \times 10 = 49 \text{ W}
\]
\[
P_{\text{hp}} = \frac{49}{746} \approx 6.6 \times 10^{-2} \text{ hp}
\]

25. An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3.00 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw, what horsepower does he use ?

Solution:

\[
W = \Delta K + \Delta U = \frac{1}{2} \times 0.2 \times 3^2 + 0.2 \times 9.8 \times 1.5 = 3.84 \text{ J}
\]
\[
P_{\text{hp}} = \frac{3.84}{746} \approx 5.14 \times 10^{-3} \text{ hp}
\]

26. In a factory it is desired to lift 2000 kg of metal through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.

Solution:

\[
P = \frac{mgh}{t} = \frac{2000 \times 9.8 \times 12}{60} = 3920 \text{ W}
\]
\[
P_{\text{hp}} = \frac{3920}{746} \approx 5.3 \text{ hp}
\]

27. A scooter company gives the following specifications… Weight — 95 kg, Max speed — 60 km/h, Max engine power — 3.5 hp, Pick up time — 5 s. Check the validity of these specifications.

Solution:

\[
v = 16.67 \text{ m/s}, \quad P = \frac{0.5 \times 95 \times (16.67)^2}{5} \approx 2639 \text{ W}
\]
Given \(3.5 \text{ hp} \approx 2611 \text{ W}\), the specifications are overclaimed.

28. A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed of 40.0 cm/s in dropping down 2.00 m, find the work done by the chain during the process.

Solution:

\[
W_{\text{chain}} + 30 \times 9.8 \times 2 = \frac{1}{2} \times 30 \times (0.4)^2
\]
\[
W_{\text{chain}} + 588 = 2.4 \Rightarrow W_{\text{chain}} \approx -586 \text{ J}
\]

29. The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

Solution:

\[
\Delta U \approx 19.6 \text{ J}
\]

30. The two blocks in an Atwood machine have masses 2.0 kg and 3.0 kg. Find the work done by gravity during the fourth second after the system is released from rest.

Solution:

\[
W_g \approx 67 \text{ J}
\]

31. Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.

Solution:

According to the figure (8-E2), the system consists of a 4.0 kg block on a horizontal table connected by a string to a 1.0 kg hanging block.

Initial state: \(u = 0\).

Final state: \(d = 1 \text{ m}, \; v = 0.3 \text{ m/s}\).

Using work-energy theorem:

Work by gravity:
\[
W_g = m g d = 1.0 \times 9.8 \times 1 = 9.8 \text{ J}
\]

Work by friction:
\[
W_f = -\mu_k (m g) d = -\mu_k \times 4.0 \times 9.8 \times 1 = -39.2\mu_k
\]

Change in kinetic energy:
\[
\Delta K = \frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}(5)(0.3)^2 = 0.225 \text{ J}
\]

Applying \(W_g + W_f = \Delta K\):

\[
9.8 – 39.2\mu_k = 0.225
\]

\[
39.2\mu_k = 9.575 \Rightarrow \mu_k \approx \mathbf{0.244}
\]

32. A block of mass 100 g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.

Solution:

\(m = 0.1 \text{ kg}, \; R = 0.1 \text{ m}, \; v_i = 5.0 \text{ m/s}\)

Initial energy:
\[
E_i = \frac{1}{2}mv_i^2 + mg(2R)
= \frac{1}{2}(0.1)(5)^2 + (0.1)(9.8)(0.2)
= 1.25 + 0.196 = 1.446 \text{ J}
\]

Final energy:
\[
E_f = 0
\]

Work done by tube:
\[
W = E_f – E_i = 0 – 1.446 = \mathbf{-1.446 \text{ J}}
\]

33. A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by the friction).

Solution:

\(m = 1400 \text{ kg}, \; v = 54 \text{ km/h} = 15 \text{ m/s}, \; h = 10 \text{ m}\)

Initial kinetic energy:
\[
K_i = \frac{1}{2}mv^2 = \frac{1}{2}(1400)(15)^2 = 157500 \text{ J}
\]

Final potential energy:
\[
U_f = mgh = 1400 \times 9.8 \times 10 = 137200 \text{ J}
\]

Using energy balance:
\[
K_i + W_f = U_f
\]

\[
157500 + W_f = 137200
\Rightarrow W_f = -20300 \text{ J}
\]

Work done against friction:
\[
\mathbf{20300 \text{ J}}
\]

34. A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline ? What will be the speed of the block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ? Take \(g = 10 \text{ m/s}^2\).

Solution:

\(m = 0.2 \text{ kg}, \; h = 3.2 \text{ m}\)

(a) Work to lift vertically:

\[
W = mgh = 0.2 \times 10 \times 3.2 = \mathbf{6.4 \text{ J}}
\]

(b) Work to slide up the incline:

\[
W = \mathbf{6.4 \text{ J}}
\]

(c) Speed if it falls vertically:

\[
mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 3.2} = \mathbf{8.0 \text{ m/s}}
\]

(d) Speed if it slides down:

\[
v = \mathbf{8.0 \text{ m/s}}
\]

35. In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m (figure 8-E3). Vertical ladder are provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three-tenth of his weight. Find (a) the work done by the boy on the ladder as he goes up, (b) the work done by the slide on the boy as he comes down, (c) the work done by the ladder on the boy as he goes up. Neglect any work done by forces inside the body of the boy.

Solution:

\(mg = 200 \text{ N}, \; h = 8.0 \text{ m}, \; L = 10 \text{ m}\)

Friction force:
\[
f = \frac{3}{10} \times 200 = 60 \text{ N}
\]

(a) Work done by the boy on the ladder:

\[
\mathbf{0}
\]

(b) Work done by the slide on the boy:

\[
W = -fL = -60 \times 10 = \mathbf{-600 \text{ J}}
\]

(c) Work done by the ladder on the boy:

\[
W = 200 \times 8 = \mathbf{1600 \text{ J}}
\]

36. Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground ?

Solution:

Height at A: \(1.0 \text{ m}\), exit height: \(0.5 \text{ m}\)

Velocity at exit:

\[
mg(1.0) = mg(0.5) + \frac{1}{2}mv^2
\Rightarrow \frac{1}{2}v^2 = 0.5g
\Rightarrow v = \sqrt{10}
\]

Time of fall:

\[
0.5 = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{0.1}
\]

Horizontal distance:

\[
x = vt = \sqrt{10} \times \sqrt{0.1} = \mathbf{1 \text{ m}}
\]

37. A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure 8-E5). The rough surface has a friction coefficient of 0.20 with the block. If the block starts slipping on the track from a point 1.0 m above the horizontal surface, how far will it move on the rough surface?

Solution:

• Initial Potential Energy ($U_i$): $mgh = 10 \text{ N} \times 1.0 \text{ m} = 10 \text{ J}$.
• Work-Energy Theorem: As the block slides down the smooth track, all its potential energy is converted into kinetic energy at the bottom ($K = 10 \text{ J}$). On the rough horizontal surface, the work done by friction ($W_f$) must equal the change in kinetic energy to bring the block to a stop.
• Frictional Force ($f$): $f = \mu \times \text{Weight} = 0.20 \times 10 \text{ N} = 2 \text{ N}$.
• Calculation:
  $W_f = \Delta K$
  $-f \times d = 0 – 10 \text{ J}$
  $-2 \text{ N} \times d = -10 \text{ J}$
  $d = \mathbf{5.0 \text{ m}}$.

38. A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.

Solution:

• Hanging Part Mass and Length: Since the chain is uniform and two-thirds is on the table, the hanging part has length $l/3$ and mass $m/3$.
• Center of Mass (CM) of Hanging Part: The CM of the hanging segment is located at its midpoint, which is $(l/3)/2 = l/6$ below the table’s surface.
• Work Done ($W$): The work required to lift the chain back onto the table is equal to the increase in the gravitational potential energy of the hanging part.
• Calculation:
  $W = \Delta U = (\text{mass of hanging part}) \times g \times (\text{height to be raised})$
  $W = (m/3) \times g \times (l/6) = \mathbf{mgl/18}$.

39. A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is $\mu$. Find the work done by the friction during the period the chain slips off the table.

Solution:

• Friction Force ($f$): At any instant, the friction force depends on the mass of the chain remaining on the table. If $x$ is the length of the chain on the table, the mass on the table is $(M/L)x$, and the friction is $f = \mu(M/L)xg$.
• Work Done ($W_f$): As the chain slips off, the length on the table changes from $x = 2L/3$ to $x = 0$. Work is the integral of the frictional force over the displacement.
• Calculation:
  $W_f = \int_{2L/3}^{0} \mu \frac{M}{L} x g \, dx = \mu \frac{Mg}{L} \left[ \frac{x^2}{2} \right]_{2L/3}^{0}$
  $W_f = \mu \frac{Mg}{L} \left[ 0 – \frac{1}{2} \left(\frac{4L^2}{9}\right) \right] = \mathbf{- \frac{2\mu MgL}{9}}$.

40. A block of mass 1 kg is placed at the point A of a rough track shown in figure (8-E6). If slightly pushed towards right, it stops at the point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.

Solution:

• Heights: From Figure 8-E6, the height at point A ($h_A$) is $1.0 \text{ m}$ and the height at point B ($h_B$) is $0.8 \text{ m}$.
• Initial and Final Kinetic Energy: Since the block is “slightly pushed” from A and stops at B, both the initial and final kinetic energies are zero.
• Work-Energy Theorem: $W_{gravity} + W_{friction} = \Delta K$
• Calculation:
  $W_{gravity} = mgh_A – mgh_B = mg(h_A – h_B)$
  $W_{gravity} = 1 \text{ kg} \times 10 \text{ m/s}^2 \times (1.0 \text{ m} – 0.8 \text{ m}) = 2 \text{ J}$
  $2 \text{ J} + W_{friction} = 0$
  $W_{friction} = \mathbf{-2 \text{ J}}$.

41. A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2.0 m/s. How high will it rise ? Take $g = 10 \text{ m/s}^2$.

Solution:

• Spring Constant ($k$): At equilibrium, the weight of the block is balanced by the spring force ($mg = kx_0$).
  $k = \frac{mg}{x_0} = \frac{5.0 \text{ kg} \times 10 \text{ m/s}^2}{0.1 \text{ m}} = 500 \text{ N/m}$.
• Total Energy at Equilibrium: Taking the equilibrium position as the reference for gravitational potential energy:
  $E_i = \text{Kinetic Energy} + \text{Elastic Potential Energy} = \frac{1}{2}mv^2 + \frac{1}{2}kx_0^2$
  $E_i = \frac{1}{2}(5.0)(2.0)^2 + \frac{1}{2}(500)(0.1)^2 = 10 \text{ J} + 2.5 \text{ J} = 12.5 \text{ J}$.
• Energy at Maximum Height ($h$): At the highest point, the velocity is zero. The spring stretch becomes $(x_0 – h)$.
  $E_f = mgh + \frac{1}{2}k(x_0 – h)^2 = (5.0)(10)h + \frac{1}{2}(500)(0.1 – h)^2$.
• Calculation:
  $12.5 = 50h + 250(0.01 – 0.2h + h^2)$
  $12.5 = 50h + 2.5 – 50h + 250h^2$
  $10 = 250h^2 \implies h^2 = \frac{1}{25} \implies h = \frac{1}{5} \text{ m} = \mathbf{20 \text{ cm}}$.

42. A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed to have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise ? Take $g = 10 \text{ m/s}^2$.

Solution:

• Initial Energy: The energy stored in the compressed spring is $U_s = \frac{1}{2}kx^2$.
  $U_s = \frac{1}{2}(100 \text{ N/m})(0.1 \text{ m})^2 = 0.5 \text{ J}$.
• Energy Conversion: When released, this elastic potential energy is converted into gravitational potential energy ($mgh$) at the block’s highest point (where kinetic energy is zero).
• Calculation:
  $mgh = 0.5 \text{ J}$
  $(0.25 \text{ kg})(10 \text{ m/s}^2)h = 0.5$
  $2.5h = 0.5 \implies h = 0.2 \text{ m} = \mathbf{20 \text{ cm}}$.

43. Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination 37°. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline. Find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take $g = 10 \text{ m/s}^2$.

Solution:

• Work-Energy for Downward Motion: The block moves $d + x = 4.8 + 0.2 = 5.0 \text{ m}$.
  $mg(d+x)\sin 37^\circ = \frac{1}{2}kx^2 + \mu(mg \cos 37^\circ)(d+x)$
  $(2)(10)(5)(0.6) = \frac{1}{2}k(0.2)^2 + \mu(2)(10)(0.8)(5)$
  $60 = 0.02k + 80\mu$ (Eq. 1).
• Work-Energy for Rebound: The block moves back $r = 1.0 \text{ m}$ from maximum compression.
  $\frac{1}{2}kx^2 = mgr\sin 37^\circ + \mu(mg \cos 37^\circ)r$
  $0.02k = (2)(10)(1)(0.6) + \mu(2)(10)(0.8)(1)$
  $0.02k = 12 + 16\mu$ (Eq. 2).
• Solving the Equations:
  (a) $60 = (12 + 16\mu) + 80\mu \implies 48 = 96\mu \implies \mathbf{\mu = 0.5}$
  (b) $0.02k = 12 + 16(0.5) = 20 \implies \mathbf{k = 1000 \text{ N/m}}$.

44. A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.

Solution:

• Initial Kinetic Energy: $K_i = \frac{1}{2}mv^2$.
• Final Kinetic Energy: $K_f = \frac{1}{2}m(v/2)^2 = \frac{1}{8}mv^2$.
• Work-Energy Theorem:
  $\frac{1}{2}mv^2 – \frac{1}{8}mv^2 = \frac{1}{2}kx^2$
  $\frac{3}{8}mv^2 = \frac{1}{2}kx^2$
  $\mathbf{k = \frac{3mv^2}{4x^2}}$.

45. Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring.

Solution:

• Conservation of Energy: In the position of maximum elongation ($x$), the block has momentarily come to rest. The gravitational potential energy lost by the hanging mass $m$ is entirely converted into the elastic potential energy of the spring.
• Calculation:
  $mgx = \frac{1}{2}kx^2$
  $mg = \frac{1}{2}kx$
  $\mathbf{x = \frac{2mg}{k}}$.

46. A block of mass $m$ is attached to two unstretched springs of spring constants $k_1$ and $k_2$ as shown in figure (8-E9). The block is displaced towards right through a distance $x$ and is released. Find the speed of the block as it passes through the mean position shown.

Solution:

When the block is displaced to the right by a distance $x$, both springs are deformed (one is stretched and the other is compressed) by the same distance $x$.

1. Initial Potential Energy: The total elastic potential energy stored in the system is the sum of the energy in both springs:
   $$U_i = \frac{1}{2}k_1 x^2 + \frac{1}{2}k_2 x^2 = \frac{1}{2}(k_1 + k_2)x^2.$$
2. Energy at Mean Position: At the mean position, the springs are unstretched ($U_f = 0$), and all the initial potential energy has been converted into the block’s kinetic energy:
   $$K_f = \frac{1}{2}mv^2.$$
3. Conservation of Energy:
   $$\frac{1}{2}(k_1 + k_2)x^2 = \frac{1}{2}mv^2.$$
4. Final Speed: Solving for $v$ gives
   $$v = \sqrt{\frac{k_1 + k_2}{m}}x.$$

47. A block of mass $m$ sliding on a smooth horizontal surface with a velocity $\vec{v}$ meets a long horizontal spring fixed at one end and having spring constant $k$ as shown in figure (8-E10). Find the maximum compression of the spring. Will the velocity of the block be the same as $\vec{v}$ when it comes back to the original position shown?

Solution:

1. Maximum Compression: At the point of maximum compression ($x_{max}$), the block momentarily stops, and its entire initial kinetic energy is stored as elastic potential energy in the spring:
   $$\frac{1}{2}mv^2 = \frac{1}{2}kx_{max}^2 \implies x_{max} = \mathbf{v\sqrt{\frac{m}{k}}}.$$
2. Velocity on Return: Since the surface is smooth (frictionless), the mechanical energy of the system is conserved. When the block returns to the original position, it will have the same speed $v$. However, velocity is a vector quantity; because the block is now moving in the opposite direction, its velocity is $-\vec{v}$. Therefore, the velocity is not the same as the original $\vec{v}$.

48. A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm (figure 8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?

Solution:

1. Speed at Release ($v$): The elastic potential energy stored in the spring is converted into kinetic energy:
   $$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$
   $$100 \times (0.05)^2 = 0.1 \times v^2$$
   $$0.25 = 0.1v^2 \implies v^2 = 2.5 \implies v = \sqrt{2.5} \text{ m/s}.$$
2. Time of Flight ($t$): The block acts as a projectile falling from a height $h = 2 \text{ m}$. Using $g = 10 \text{ m/s}^2$:
   $$h = \frac{1}{2}gt^2$$
   $$2 = \frac{1}{2}(10)t^2 \implies t^2 = 0.4 \implies t = \sqrt{0.4} \text{ s}.$$
3. Horizontal Distance ($D$):
   $$D = v \times t = \sqrt{2.5} \times \sqrt{0.4} = \sqrt{1} = \mathbf{1 \text{ m}}.$$
   The block will hit the ground 1 m from the free end of the spring.

49. A small heavy block is attached to the lower end of a light rod of length $l$ which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle?

Solution:

For a particle attached to a rigid rod (unlike a flexible string), the minimum condition to complete a vertical circle is that the particle must just reach the highest point with zero velocity ($v_{top} \ge 0$).

1. Conservation of Energy:
   $$E_{bottom} = \frac{1}{2}mv_0^2$$
   $$E_{top} = mg(2l) + \frac{1}{2}mv_{top}^2$$
2. Calculation: Setting $v_{top} = 0$:
   $$\frac{1}{2}mv_0^2 = 2mgl \implies v_0^2 = 4gl$$
   $$v_0 = \mathbf{2\sqrt{gl}}.$$

50. Figure (8-E12) shows two blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take $g = 10 \text{ m/s}^2$.

Solution:

1. Break-off Condition: Block A leaves the surface when the vertical component of the spring force equals its weight ($mg = 0.32 \times 10 = 3.2 \text{ N}$).
   Let $L$ be the spring length at break-off. The vertical component is:
   $$k(L – 0.4)\cdot \left(\frac{0.4}{L}\right) = 3.2.$$
   Solving:
   $$16\left(1 – \frac{0.4}{L}\right) = 3.2 \implies L = 0.5 \text{ m}.$$
2. Displacement: The horizontal distance moved by A (and thus the vertical distance dropped by B) is:
   $$x = \sqrt{0.5^2 – 0.4^2} = 0.3 \text{ m}.$$
   The spring extension is:
   $$\Delta L = 0.5 – 0.4 = 0.1 \text{ m}.$$
3. Conservation of Energy:
   $$m_B gx = \frac{1}{2}k(\Delta L)^2 + \frac{1}{2}(m_A + m_B)v^2$$
   $$0.32 \times 10 \times 0.3 = \frac{1}{2}(40)(0.1)^2 + \frac{1}{2}(0.64)v^2$$
   $$0.96 = 0.2 + 0.32v^2 \implies 0.76 = 0.32v^2 \implies v = \mathbf{1.5 \text{ m/s}}.$$

51. One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (figure 8-E13). Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.

Solution:

• Initial State: Length $L_i = \frac{h}{\cos 37^\circ} = \frac{h}{0.8} = 1.25h$. Initial extension $x_i = 0.25h = \frac{h}{4}$.
• Final State: Spring is vertical, length $L_f = h$, extension $x_f = 0$.
• Energy Conservation:
  $$\frac{1}{2} k x_i^2 = \frac{1}{2} mv^2 \implies k\left(\frac{h}{4}\right)^2 = mv^2 \implies v^2 = \frac{kh^2}{16m}.$$
  $$v = \mathbf{\frac{h}{4} \sqrt{\frac{k}{m}}}.$$

52. Figure (8-E14) shows a light rod of length l rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth h below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the block moves in a complete circle about the ring ?

Solution:

• The block reaches the ring level with speed $v = \sqrt{2gh}$. It is at distance $l$ from the ring.
• For a mass attached to a rigid rod, the minimum speed required at the bottom to complete a vertical circle is that which allows it to just reach the top ($v_{top} \ge 0$).
• By conservation of energy:
  $$\frac{1}{2} mv^2 = mg(l) + \frac{1}{2} m v_{top}^2.$$
  Taking $v_{top} = 0$:
  $$\frac{1}{2} m (2gh) = mgl \implies gh = gl \implies \mathbf{h = l}.$$

53. The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $\sqrt{10 gl}$, where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.

Solution:

• Initial Velocity: $v_0 = \sqrt{10gl}$.
• (a) Horizontal ($\phi = 90^\circ$): Height $h = l$.
  $$v^2 = v_0^2 – 2gl = 8gl.$$
  $$T = \frac{mv^2}{l} + mg \cos 90^\circ = \frac{m(8gl)}{l} = \mathbf{8mg}.$$
• (b) Highest point ($\phi = 180^\circ$): Height $h = 2l$.
  $$v^2 = v_0^2 – 4gl = 6gl.$$
  $$T = \frac{mv^2}{l} + mg \cos 180^\circ = \frac{m(6gl)}{l} – mg = \mathbf{5mg}.$$
• (c) 60° with upward vertical ($\phi = 120^\circ$): Height $h = l(1 – \cos 120^\circ) = 1.5l$.
  $$v^2 = v_0^2 – 2g(1.5l) = 7gl.$$
  $$T = \frac{mv^2}{l} + mg \cos 120^\circ = \frac{m(7gl)}{l} – 0.5mg = \mathbf{6.5mg}.$$

54. A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

Solution:

• Parameters: $l = 0.5 \text{ m}$, $m = 0.1 \text{ kg}$, $h = l(1 – \cos 37^\circ) = 0.5(1 – 0.8) = 0.1 \text{ m}$.
• Velocity at bottom:
  $$v^2 = 2gh = 2 \times 9.8 \times 0.1 = 1.96.$$
• Tension:
  $$T = \frac{mv^2}{l} + mg = \frac{0.1 \times 1.96}{0.5} + 0.98 = 1.372 \approx \mathbf{1.4 \text{ N}}.$$

55. Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.

Solution:

• At point P (horizontal radius), the normal force $N = \frac{mv^2}{R}$.
• Given $N = mg \implies v^2 = Rg$.
• Energy Conservation:
  $$\frac{1}{2} k x^2 = mgR + \frac{1}{2} m v^2.$$
  $$\frac{1}{2} k x^2 = mgR + \frac{1}{2} mgR = \frac{3}{2} mgR.$$
  $$x^2 = \frac{3mgR}{k} \implies \mathbf{x = \sqrt{\frac{3mgR}{k}}}.$$

60. A particle of mass \(m\) is kept on a fixed, smooth sphere of radius \(R\) at a position, where the radius through the particle makes an angle of \(30^\circ\) with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release ? (b) Find the distance travelled by the particle before it leaves contact with the sphere.

Solution:

(a) Force exerted just after release:

At the instant of release, \(v = 0\).

Radial equation:
\[
mg\cos 30^\circ – N = \frac{mv^2}{R}
\]

Since \(v = 0\):

\[
N = mg\cos 30^\circ = mg \cdot \frac{\sqrt{3}}{2}
\]

\[
\mathbf{N = \frac{\sqrt{3}}{2}mg}
\]

(b) Distance travelled before leaving contact:

Condition for leaving contact (\(N = 0\)):

\[
mg\cos\theta = \frac{mv^2}{R}
\Rightarrow v^2 = Rg\cos\theta
\]

Using conservation of energy:

\[
mgR\cos 30^\circ = mgR\cos\theta + \frac{1}{2}mv^2
\]

Substitute \(v^2 = Rg\cos\theta\):

\[
mgR\frac{\sqrt{3}}{2} = mgR\cos\theta + \frac{1}{2}m(Rg\cos\theta)
\]

\[
\frac{\sqrt{3}}{2} = \cos\theta + \frac{1}{2}\cos\theta = \frac{3}{2}\cos\theta
\]

\[
\cos\theta = \frac{\sqrt{3}}{3}
\]

Angle covered:

\[
\Delta \theta = \cos^{-1}\left(\frac{\sqrt{3}}{3}\right) – 30^\circ \approx 24.7^\circ
\]

Convert to radians:

\[
\Delta \theta \approx 0.43 \text{ rad}
\]

Distance travelled:

\[
s = R \Delta \theta = \mathbf{0.43R}
\]

61. A particle of mass \(m\) is kept on the top of a smooth sphere of radius \(R\). It is given a sharp impulse which imparts it a horizontal speed \(v\). (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of \(v\) for which the particle does not slip on the sphere? (c) Assuming the velocity \(v\) to be half the minimum calculated in part (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Solution:

(a) At the topmost point:

\[
mg – N = \frac{mv^2}{R}
\]

\[
\mathbf{N = mg – \frac{mv^2}{R}}
\]

(b) For the particle to just lose contact at the top, \(N = 0\):

\[
mg = \frac{mv^2}{R} \Rightarrow v = \mathbf{\sqrt{Rg}}
\]

(c) Given \(v = \frac{1}{2}\sqrt{Rg}\)

At the point of leaving contact, \(N = 0\):

\[
mg\cos\theta = \frac{mv_\theta^2}{R}
\Rightarrow v_\theta^2 = Rg\cos\theta
\]

Using conservation of energy:

\[
mgR + \frac{1}{2}m\left(\frac{\sqrt{Rg}}{2}\right)^2 = mgR\cos\theta + \frac{1}{2}mv_\theta^2
\]

\[
mgR + \frac{1}{8}mgR = mgR\cos\theta + \frac{1}{2}m(Rg\cos\theta)
\]

\[
\frac{9}{8}mgR = \frac{3}{2}mgR\cos\theta
\]

\[
\cos\theta = \frac{3}{4}
\Rightarrow \mathbf{\theta = \cos^{-1}\left(\frac{3}{4}\right)}
\]

62. Figure (8-E17) shows a smooth track which consists of a straight inclined part of length \(l\) joining smoothly with the circular part. A particle of mass \(m\) is projected up the incline from its bottom. (a) Find the minimum projection-speed \(v_0\) for which the particle reaches the top of the track. (b) Assuming that the projection-speed is \(2v_0\) and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than \(v_0\), where will the block lose contact with the track?

Solution:

(a) Total height:

\[
H = l\sin\theta + R(1 – \cos\theta)
\]

Using energy conservation:

\[
\frac{1}{2}mv_0^2 = mgH
\]

\[
\mathbf{v_0 = \sqrt{2g\left[l\sin\theta + R(1 – \cos\theta)\right]}}
\]

(b) If \(v = 2v_0\):

\[
K = \frac{1}{2}m(2v_0)^2 = 4mgH
\]

At the top:

\[
K_{\text{top}} = 4mgH – mgH = 3mgH
\]

\[
\frac{1}{2}mv^2 = 3mgH \Rightarrow \frac{mv^2}{R} = \frac{6mgH}{R}
\]

Radial equation:

\[
N + mg\cos\theta = \frac{mv^2}{R}
\]

\[
\mathbf{N = \frac{6mg\left[l\sin\theta + R(1 – \cos\theta)\right]}{R} – mg\cos\theta}
\]

(c) The block loses contact when \(N = 0\). This occurs at:

\[
\mathbf{\theta = \cos^{-1}\left(\frac{2}{3}\right)}
\]

63. A chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R > l\) with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle \(\theta\). (c) Find the tangential acceleration \(dv/dt\) of the chain when the chain starts sliding down.

Solution:

(a) Gravitational potential energy:

\[
U = \int_0^{l/R} \frac{mR^2g}{l} \cos\phi \, d\phi
= \frac{mR^2g}{l} \left[\sin\phi\right]_0^{l/R}
\]

\[
\mathbf{U = \frac{mR^2g}{l} \sin\left(\frac{l}{R}\right)}
\]

(b) Kinetic energy after sliding through angle \(\theta\):

New potential energy:

\[
U’ = \frac{mR^2g}{l} \left[\sin\left(\theta + \frac{l}{R}\right) – \sin\theta\right]
\]

Using \(K = U – U’\):

\[
\mathbf{K = \frac{mR^2g}{l} \left[ \sin\left(\frac{l}{R}\right) + \sin\theta – \sin\left(\theta + \frac{l}{R}\right) \right]}
\]

(c) Tangential acceleration at start (\(\theta = 0\)):

\[
a = \frac{Rg}{l}\left[\cos\theta – \cos\left(\theta + \frac{l}{R}\right)\right]
\]

At \(\theta = 0\):

\[
\mathbf{a = \frac{Rg}{l}\left[1 – \cos\left(\frac{l}{R}\right)\right]}
\]

64. A smooth sphere of radius \(R\) is made to translate in a straight line with a constant acceleration \(a\). A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle \(\theta\) it slides.

Solution:

Using work-energy theorem in accelerating frame:

\[
\frac{1}{2}mv^2 = mgR(1 – \cos\theta) + maR\sin\theta
\]

\[
v^2 = 2R\left[g(1 – \cos\theta) + a\sin\theta\right]
\]

\[
\mathbf{v = \sqrt{2R\left(g(1 – \cos\theta) + a\sin\theta\right)}}
\]