HC Verma Concepts Of Physics Exercise Solutions Of Chapter 8 (Work and Energy)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 8 - Work and Energy:

EXERCISE

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.

Sol:

Given: total mass, M = 90 kg; initial speed, u = 6 km/h = 5/3 m/s; final velocity, v = 12 km/h = 10/3 m/s.

Increase in kinetic energy,

→ ΔK.E. = ½ M (v² – u²)

Or, ΔK.E. = ½ * 90 * [(10/3)² – (5/3)²]

Or, ΔK.E. = 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.

Sol:

Given: mass, m = 2 kg; initial speed, u = 10 m/s; acceleration, a = 3 m/s²; time, t = 5 s.

→ Final velocity, v = u + at

Or, v = 10 + 3 * 5

Or, v = 25 m/s

Therefore, Final kinetic energy = ½ mv²

Or, Final kinetic energy = ½ * 2 * 25

Or, Final kinetic energy = 25 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?

Sol:

Given: resisting force, F = 100 N; displacement, s = 4 m.

→ Work is done, w = F * s

Or, w = 100 * 4 = 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.

Sol:

Given: mass, m = 5 kg; length of inclination, l = 10 m; angle of inclination, θ = 30⁰; g = 9.8 m/s².

Force along inclination, F = mg sin θ

Or, F = 5 * 9.8 * sin 30⁰

Or, F = 24.5 N

So, Work done by the force of gravity = F * l = 24.5 * 10 = 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.

Sol:

Given: force, F = 2.5 N; displacement, s = 2.5 m; mass, m = 15 g = 0.015 kg; initial velocity, u = 0; a = 2.5/0.015 = 500/3 m/s².

The work done, w = F * s

Or, w = 2.5 * 2.5 = 6.25 J.

Applying work-energy principle,

→ ½ mv² – ½ mu² = w

Or, ½ * 0.015 * v² – 0 = 6.25

Or, v = 28.86 m/s

We know, t = (v – u)/a

Or, t = (28.86 – 0) * 3/500

Or, t = 0.173 s.

So, time taken to travel this distance is 0.173 s.

Therefore, average power delivered is = w/t = 36.1 watt.

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.

Sol:

Displacement, r = r₁ – r₂

Or, r = [(2 m) i + (3 m) j] – [(3 m) i + (2 m) j]

Or, r = (– 1 m) i + (1 m) j

The work done, w = F. r

Or, w = [(5 N) i + (5 N) j]. [(– 1 m) i + (1 m) j]

Or, w = – 5 + 5 = 0

So, the work done by the force on the particle during the displacement is zero.

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.

Sol:

Given: mass of block, m = 2 kg; acceleration, a = 0.5 m/s²; distance covered, s = 40 m.

Work done, w = F * s

Or, w = m * a * s

Or, w = 2 * 0.5 * 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.

Sol:                                     

Work done, dw = F.dx

Or, w = ∫dw = ∫F.dx

Or, w = ∫ (a + bx) dx

Limit of integration: from 0 to d.

Or, w = [ax + bx²/2]^d_o 

Or, w = [ad + bd²/2]

Or, w = d [a + bd/2].

9. A block of mass 250 g slides down an incline of inclination 37⁰ with a uniform speed. Find the work done against the friction as the block slides through 1.0 m. 

Sol:

Given: mass of block, m = 250 g = 0.25 kg; θ = 37⁰; displacement, s = 1 m.

For uniform velocity, v

Frictional force, f = mg sin θ

Or, f = 0.25 * 9.8 * sin 37⁰

Or, f = 1.47 N

Work done due to frictional force, w = f * s = 1.47 * 1 = 1.47 J.

10. A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (figure 8-E1). 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, (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface, (b) Find the frictional force acting on the smaller block, (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system.

Sol:

Given: acceleration, a = F/2(m + M).

From the F.B.D.

The equation of motion of block of mass ‘m’

Vertical motion, R₁ = mg

Horizontal motion, ma = f₁

Or, f₁ = mF/2(m + M)

The equation of motion of block of mass ‘M’

 Vertical motion, R₂ = Mg + R₁ = (m + M) g

Horizontal motion, Ma = F – f₁ – f₂

Or, MF/2(m + M) = F – mF/2(m + M) – μ₂R₂

Or, μ₂ (m + M) g = F – F/2

Or, μ₂ = F/ [2(m + M) g]

(a) The coefficient of kinetic friction between the bigger block and the horizontal surface is F/ [2(m + M) g].

(b) The frictional force acting on the smaller block is mF/2(m + M).

(c) The work done, w = 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.

Sol:

Given: weight of the box, mg = 2000 N; μ = 0.2; distance travelled, s = 20 m.

(a) Vertical motion:

→ R + P sin θ – mg = 0

Or, R = 2000 – P sin θ

Horizontal motion:

→ ma = P cos θ – f

Or, P cos θ – μR = 0     (a = 0, because box slid slowly)

Or, P cos θ – 0.2 (2000 – P sin θ) = 0

Or, P = 400/ (cos θ + 0.2 sin θ)

Therefore, the work done by the person W = P * s* cos θ

Or, W = 40000 / (5 + tan θ).

(b) For the minimum magnitude of force, P_min

→ d (cos θ + 0.2 sin θ)/dθ = 0

Or, – sin θ + 0.2 cos θ = 0

Or, tan θ = 0.2

Therefore, work done W = 40000 / (5 + tan θ)

Or, W = 40000 / (5 + 0.2) = 7692 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.

Sol:

(a) Motion in y - direction:

→ R – 100 cos 37 = 0

Motion in x - direction:

→ Ma = P – 100 sin 37

Acceleration, a = 0, because block is sliding slowly.

∴ 0 = P – 100 sin 37

Or, P = 100 sin 37

The work done by the person in moving the block through a distance of 2.0 m is = P * s = (100 sin 37) * 2 = 120 J.

(b) Motion in y - direction:

→ R – P sin 37 – 100 cos 37 = 0

Motion in x - direction:

→ Ma = P cos 37 – 100 sin 37

Acceleration, a = 0, because block is sliding slowly.

∴ 0 = P cos 37 – 100 sin 37

Or, P = 100 tan 37

The work done by the person in moving the block through a distance of 2.0 m is = (P cos 37) * s = (100 sin 37) * 2 = 120 J.

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.

Sol:

Given: initial speed, u = 72 km/h = 20 m/s; distance travelled, s = 25 m; final speed, v = 0, m = 500 kg; frictional force, f =?

We know, v² – u² = 2as

Or, 0² – 20² = 2 * a * 25

Or, a = – 8 m/s² (–ve sign means accel. Is opposite to displacement)

Frictional force, f = ma = 500 * 8 = 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.

Sol:

Given: mass, m = 500 kg; initial speed, u = 0 m/s; final speed, v = 72 km/h = 20 m/s; s = 25 m.

We know, v² – u² = 2as

Or, 20² – 0² = 2 * a * 25

Or, a = 8 m/s² 

Therefore, force F = ma = 500 * 8 = 4000 N.

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

Sol:

Given: v = α√x; mass = m; distance, s = d.

Acceleration, a = dv/dt

Or, a = (α/√x) * dx/dt = (α/2√x) * v

Or, a = (α/2√x) * α√x = α²/2

Force, F = ma = m α²/2

Therefore, the total work done W = F * s = mdα²/2.

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 m/s².

Sol:

Given: mass, m = 2 kg; Force, F = 20 N acts for one second.

Initial velocity, u = 0; t = 1 s.

Motion of the block along the plane

→ Ma = F – mg sin 37

 Or, 2a = 20 – 12 → a = 4 m/s²

Distance travelled, s = ut + ½ at² → s = 2 m.

(a) The work done by the applied force = F * s = 20 * 2 = 40 J.

(b) The work done by the force of gravity = – mg sin 37 * s

= – 2 * 10 * sin 37 * 2 = – 24 J.

Here, –ve sign shows that gravity force acts opposite to the displacement.

(c) Velocity of the block at the instant the force ceases to act is, v = u + at = 0 + 4 * 1 = 4 m/s.

∴ The kinetic energy of the block, KE = ½ mv² = ½ * 2 * 4² = 16 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 m/s².

Sol:

Given: Force, F = 20 N; mass, m = 2 kg; acceleration, a = 10 m/s²; initial velocity, u = 0.

(a) Time, t = 1 s.

Distance travelled, s = ut + ½ at

Or, S = 0 + ½ * 10 * 1 = 5 m.

Therefore, the work done by the applied force in the first second, W = F * s = 20 * 5 = 100 J.

(b) The work done by the force of gravity = mg sin 37 * s

= 2 * 10 * sin 37 * 5 = 60 J.

Here, +ve sign shows that gravity force and displacement acts in same direction.

(c) Motion of the block along the plane

→ ma = F + mg sin 37 – F_f

Or, Frictional force, F_f = F + mg sin 37 – ma

Or, F_f = 20 + 12 – 20 = 12 N.

Therefore, the work done by the frictional force = – F_f * s = – 12 * 5 = – 60 J.

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?

Sol:

Given: mass, m = 250 g = 0.25 kg; initial velocity, u = 40 cm/s = 0.4 m/s; final velocity, v = 0, μ = 0.1.

Frictional force, F_f = μR

Or, F_f = μmg = 0.1 * 0.25 * 9.8 = 0.245 N.

Acceleration, a = – F_f/m = – 0.98 m/s²

Distance travelled, s = (v² – u²)/2a = 0.082 m = 8.2 cm.

Therefore, work done, W = – F_f * s = – 0.245 * 0.082 = – 0.02 J.

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?

Sol:

Given: height, H = 50 m; mass, m = 1.8 x 10⁵ kg/hr = 50 kg/s.

Therefore, the gravitational potential energy per second is = mgH

PE = 50 * 9.8 * 50 = 24500 W

Electrical energy = ½ P.E = 12250 W

Therefore, number of 100 W lamps, that can be lit = 12250/100 = 122.5 ≈ 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?

Sol:

Given, mass, m = 6 kg; height, h = 2 m.

→ P.E. at a height ‘2 m’ = mgh

Or, P.E. at a height ‘2 m’ = 6 * 9.8 * 2

Or, P.E. at a height ‘2 m’ = 117.6 J

P.E. at floor = 0

Loss in P.E. = 117.6 – 0 = 117. 6 J ≈ 118 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.

Sol:

Given: h = 40 m; u = 50 m/s.

Let the speed be ‘v’ when it strikes the ground.

Applying law of conservation of energy

We get, mgh + ½ mu² = ½ mv²

Or, gh + ½ u² = ½ v²

Or, 10 × 40 + (1/2) × 2500 = ½ v²

Or, v² = 3300

Or, v = 57.4 m/s ≈ 58 m/s.

22. The 200 m free style women's swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when 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.

Sol:

Given: time, t = 1 min 57.56 sec = 117.56 s; Power, P = 460 W; distance travelled, s = 200 m.

We know, P = Work/time

Or, work, W = 460 * 117.56 J

Again, W = F * s

Or, F = (460 * 117.56)/200 = 270.3 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?

Sol:

Given: mass, m = 50 kg; time taken, t = 10.54 s; distance travelled, s = 100 m.

The motion can be assumed to be uniform because she achieved her maximum speed in a very short time.

(a) Speed of Griffith-Joyner is, v = (distance travelled/time taken)

Or, v = 100/10.54 = 9.488 m/s

Therefore, the kinetic energy of Griffith-Joyner is, KE = ½ mv²

Or, KE = ½ * 50 * (9.488)² = 2250 J.

(b) Average resistance force, F_r = weight/10 = (50*9.8)/10 = 49 N.

So, the work done by the resistance is, W = – F_r * s

Or, W = – 49 * 100 = – 4900 J.

–ve sign shows that, force is acting opposite to the displacement.

(c) To maintain her uniform speed, she has to exert 4900 J of energy to overcome friction.

Therefore, Power = work/time = 4900/10.54 = 465 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.

Sol:

Given: mass flow rate, m = 30 kg/min = 0.5 kg/s; height, h = 10 m.

So, Power = mgh

Or, Power = 0.5 * 9.8 * 10

Or, Power = 49 Watt

Or, Power = 49/746 hp = 6.6 * 10^-3 hp.

[Since, 1 hp = 746 Watt]

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?

Sol:

Given: mass, m = 200 g = 0.2 kg; height, h = 150 cm = 1.5 m; Speed, v = 3 m/s; time taken, t = 1 s.

Total work done, W = ½ mv² + mgh

Or, W = ½ * (0.2) * 9 + (0.2) * (9.8) * (1.5)

Or, W = 3.84 J.

And Power, P = W/t = 3.84/1

Or, P = 3.84 Watt

Or, P = 3.84/746 hp = 5.14 * 10^-3 hp.

[Since, 1 hp = 746 Watt]

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.

Sol:

Given: mass, m = 2000 Kg; height, h = 12 m; time taken, t = 1 min = 60 s.

So, work done by engine, W = mgh

Or, W = 2000 * 10 * 12 = 240000 J.

And, Power, P = W/t

Or, P = 240000/60

Or, P = 4000 watt = 4000/746 = 5.3 hp.

[Since, 1 hp = 746 Watt]

27. A scooter company gives the following specifications about its product.

Weight of the scooter — 95 kg

Maximum speed — 60 km/h

Maximum engine power — 3.5 hp

Pick up time to get the maximum speed — 5 s

Check the validity of these specifications.

Sol:

The specification given by the company are:

Initial velocity, u = 0; mass, m = 95 kg; Maximum engine power, P_m = 3.5 hp; maximum speed, V_m = 60 km/h = 50/3 m/s; t_m = 5 s.

So, the maximum acceleration that can be produced is given by,

a = {(50 / 3) – 0}/5 = 10/3 m/s².

So, the driving force is given by

F = ma = 95 * 10/3 = 950/3 N

So, the velocity that can be attained by maximum h.p. will be v = P/F = (3.5 * 746 * 3)/950 = 8.2 m/s.

Because, the scooter can reach a maximum of 8.2 m/s while producing a force of 950/3 N, the specifications given are somewhat over claimed.

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.

Sol:

Given: mass, m = 30 kg; final velocity, v = 40 cm/s = 0.4 m/s; initial velocity, u = 0; distance travelled, s = 2 m.

Let, T be the force applied by the chain.

We know, 2as = v² – u²

Or, a = (v² – u²)/2s

Or, a = 0.16/4 = 0.04 m/s².

Acceleration of the block, a = 0.04 m/s².

From the F.B.D. of the block,

We get, ma = mg – T

Or, T = m (g – a)

Or, T = 30 * (9.8 – 0.04) = 292.8 N.

Therefore, the work done by the chain is W = – Ts = – 292.8 * 2 = – 585.6 J. [–ve sign shows that, force and displacement in opposite direction]

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.

Sol:

Given: T = 16 N; m₁ = m; m₂ = 2m; initial velocity, u = 0; t = 1 s.

Equation of motion of block m₁:

→ ma = T – mg  --------- (1)

Equation of motion of block m₂:

→ 2ma = 2mg – T --------- (2)

Solving equation 1 and 2, we get

Acceleration, a = g/3 and mass, m = 12/g

After 1s m₁ will move upward and m₂ move downward by h.

And, h = ½ at² = g/6.

Potential energy change for m₁ = + mgh = (12/g) * g * (g/6) = + 2g and Potential energy change for m₂ = – 2mgh = 2*(12/g) * g * (g/6) = – 4g.

Therefore, the gravitational potential energy of the system is = + 2g – 4g = – 2g = – 2 * 9.8 = – 19.6 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.

Sol:

Given: m₁ = 2 kg; m₂ = 3 kg; initial speed, u = 0; n = during 4^th second.

Equation of motion of block m₁:

→ m₁a = T – m₁g 

Or, 2a = T – 2g --------- (1)

Equation of motion of block m₂:

→ m₂a = m₂g – T

Or, 3a = 3g – T --------- (2)

Solving equation 1 and 2, we get

Acceleration, a = g/5 m/s².

Distance travelled in n^th sec is given by

→ S_nth = u + ½ a (2n – 1)

Or, S_4th = 0 + ½ * (g/5) * (2 * 4 – 1)

Or, S_4th = 7g/10.

Potential energy change for m₁ = + m₁gh = 2 * g * (7g/10) = + 7g²/5 and Potential energy change for m₂ = – m₂gh = 3 * g * (7g/10) = – 21g²/10.

Therefore, the gravitational potential energy of the system is = + 7g²/5 – 21g²/10 = – 7g²/10 = – 67 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.

Sol:

Given: initial velocity, u = 0; distance travelled by 1 kg block, s = 1 m; velocity after 1 m descend, v = 0.3 m/s.

We have, 2as = v² – u²

Or, 2 * a * 1 = 0.3² – 0²

Or, a = 0.045 m/s².

Equation of motion of 4 kg block:

→ m * 2a = T – µR

Or, 4 * 2 * 0.045 = T – 4µg

Or, 0.36 = T – 4µg ------ (1)

Equation of motion of 1 kg block:

→ ma = mg – T’

Or, 1 * 0.045 = 1g – 2T

Or, 0.045 = g – 2T --------- (2)

Solving equation 1 and 2, we get

→ µ = 0.12.

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.

Sol:

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