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.âˆ†x.

To calculate the work done by a variable force integration is necessary. If we are calculating the work done by spring we apply Hookeâ€™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 (k1ï¼žk2). They are stretched by the same force then**

- More work is done in the first spring
- In both springs equal work is done
- In the second spring, more work is done
- 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 ofÂ 0.20 m has a force constant 10 N/m. When the stretch is changed to 0.25 m, the increase in potential energy isÂ **

- 0.2 joule
- 0.3 joule
- 0.1 joule
- 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?**

- 1:4
- 2:1
- 4:1
- 1:2

**Answer: (b) 2:1**

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

- 68 J
- 23 J
- 84 J
- 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?**

- 0.1 m
- 0.2 m
- 0.01 m
- 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**

- 16 J
- 8 J
- 32 J
- 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 **

- 100 J
- 200 J
- 300 J
- 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 **

- 4 J
- 54 J
- 415 J
- 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/sÂ collides with a nearly weightless spring of force constant k= 5 N/m.Â **

- 0.5 m
- 0.12 m
- 1.5 m
- 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?**

- 4 U
- 8 U
- 16 U
- 2 U

**Answer: (c) 16 U**

**More NEET Physics topics:**

- MCQs on Electric Charges and Fields
- MCQs on Photodiode
- MCQs on Work, Energy and Power
- Motion in a Straight Line MCQs
- Physical World And Measurement MCQs