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NCERT Exemplar · Class 11 · Physics

Chapter 6: Work, Energy and Power

Practice each question, then open the answer or worked solution.

Questions

48 questions
Q6.41Written answer

An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 km/h when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of the energy of the wagon is lost due to friction, calculate the spring constant.

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Worked solution

KE = 1/2 mv2

m = 50000 kg

v = 10 m/s

KE = 2500000J

KE of spring = 10% of the KE wagon

K = 5 × 105 N/m

Q6.42Written answer

An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilized by him in jogging, assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilized for jogging.

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Worked solution

The energy used up is given as = mgh

mg = 600 N

h = 0.25m

No.of steps in 6 km = 6000 steps

Energy used in 6000 m = (6000)(600)(0.25)J

Energy utilized in jogging = 9 × 104 J

Q6.43Written answer

On complete combustion, a litre of petrol gives off heat equivalent to 3 × 107 J. In a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving with a uniform speed on a surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.

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Worked solution

Efficiency of the car engine = 0.5

Energy given by the car with 1 litre of petrol = 1.5 × 107

WD = 1.5 × 107

f = 103 N

Long Answers
Q6.44Written answer

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30o by a force of 10 N parallel to the inclined surface. The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate

a) work done against gravity

b) work done against the force of friction

c) increase in potential energy

d) increase in kinetic energy

e) work done by an applied force

Exemplar Solutions Physics Class 11 Chapter 6 - 27

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Worked solution

a) Work done against gravity = mgh

h = 5 m

WD against gravity = 50 J

b) Work done against the force of friction is μfs = 5√3 J

c) Increase in PE = WD against the gravity = 50 J

d) Increase in KE is equal to the WD by the system = ∆K = WD = 41.340 J

e) WD by the applied force = Fs = 100J

Q6.45Written answer

A curved surface as shown in the figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from one by one from A, which is at a slightly greater height than C. with the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.

a) for which balls is total mechanical energy conserved?

b) which ball can reach D?

c) for balls which do not reach D, which of the balls can reach back A?

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Worked solution

a) For ball 1 the total mechanical energy is conserved

b) Ball 1 reaches D

c) Ball 3 reaches back A

Q6.46Written answer

A rocket accelerates straight up by ejecting gas downwards. In a short time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = (1/2) ∆m u2 in this time interval.

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Worked solution

M is the mass of the rocket at any time t

v is the velocity of the rocket

∆m is the mass of the gas that is ejected during the time interval ∆t

Therefore, K = 1/2 u2∆m

Q6.47Written answer

Two identical steel cubes collide head-on face to face with a speed of 10 cm/s each. Find the maximum compression of each. Young’s modulus for steel = Y = 2 × 1011 N/m2.

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Worked solution

Y = stress/strain

Y = FL/A∆L

WD = F∆L

KE = 5 × 10-4 J

WD = KE

∆L = 5 × 10-7 m

Q6.48Written answer

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect the viscous drag of air and assume that the density of air is constant.

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Worked solution

When the dragging viscous force of the air on the balloon is neglected, then the net buoyant force = vpg

Where v is the volume of air displaced

p is the net density upward

After calculation, it can be said that as the balloon goes up, the sum of kinetic energy and the potential energy of the balloon will be equal to change in potential energy in the air.

Important Topics of Chapter 6 Work, Energy and Power

  • Notions of work and kinetic energy: The work-energy theorem
  • Work
  • Kinetic energy
  • Work done by a variable force
  • The work-energy theorem for a variable force
  • The concept of potential energy
  • The conservation of mechanical energy
  • The potential energy of a spring
  • Forms of energy: the law of conservation of energy
  • Power
  • Collisions