Work Done by Variable Force MCQs for NEET

Work is done by a force if it acts on a body and there is a displacement of the body in the direction of the force. The work done by a constant force is given by W = F.∆x.

To calculate the work done by a variable force integration is necessary. If we are calculating the work done by spring we apply Hooke’s law to find the value of force and integrate it to find the work done.

Q1: There are two springs with the force constant as k1 and k2 (k1>k2). They are stretched by the same force then

  1. More work is done in the first spring
  2. In both springs equal work is done
  3. In the second spring, more work is done
  4. No work is done in both the springs

Answer: (c) More work is done in case of the second spring

Q2: A spring with an initial stretch of  0.20 m has a force constant 10 N/m. When the stretch is changed to 0.25 m, the increase in potential energy is 

  1. 0.2 joule
  2. 0.3 joule
  3. 0.1 joule
  4. 0.5 joule

Answer: (c) 0.1 joule

Q3: Same force is used to stretch two springs of spring constants 1500 N/m and 3000 N/m respectively. What will be the ratio of potential energy?

  1. 1:4
  2. 2:1
  3. 4:1
  4. 1:2

Answer: (b) 2:1

Q4: What will be the work done in stretching the  spring through 40 mm if 10 N force is required to stretch the spring through 1 mm

  1. 68 J
  2. 23 J
  3. 84 J
  4. 8 J

Answer: (d) 8 J

Q5: A body moving with a velocity 10 m/s and having a mass 0.1 kg hits a spring (fixed at the other end) of force constant 1000 N/m and comes to rest after compressing the spring. What will be the value of compression of the spring?

  1. 0.1 m
  2. 0.2 m
  3. 0.01 m
  4. 0.5 m

Answer: (b) 0.1 m

Q6: A spring having an extension of 5 cm has a force constant 800 N/m. The work done in extending it from 5 cm to 15 cm is

  1. 16 J
  2. 8 J
  3. 32 J
  4. 24 J

Answer: (b) 8 J

Q7: 100 J of energy is stored by a spring when it is stretched by 2 cm. If it is stretched further by 2 cm, the stored energy will be increased by

  1. 100 J
  2. 200 J
  3. 300 J
  4. 400 J

Answer: (c) 300 J

Q8: The potential energy is 4 J for spring when it is stretched by 2 mm. If it is stretched by 10 mm, its potential energy is equal to

  1. 4 J
  2. 54 J
  3. 415 J
  4. 100 J

Answer: (d) 100 J

Q9: What would be the maximum compression of the spring if a mass of 0.5 kg moving on a horizontal smooth surface with a speed of 1.5 m/s  collides with a nearly weightless spring of force constant k= 5 N/m. 

  1. 0.5 m
  2. 0.12 m
  3. 1.5 m
  4. 0.15 m

Answer: (c) 0.15 m

Q10: A spring has a spring constant k. When the spring is stretched through 1 cm, the potential energy will be U. What will be the potential energy if it is stretched by 4 cm?

  1. 4 U
  2. 8 U
  3. 16 U
  4. 2 U

Answer: (c) 16 U

 

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