# Work Done: Spring

## Trending Questions

**Q.**

A long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be

**Q.**

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

16

*J*8

*J*24

*J*32

*J*

**Q.**

The potential energy of a certain spring when stretched through a distance 'S' is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance 'S' will be

30

40

10

20

**Q.**

A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is

68

*J*84

*J*23

*J*8

*J*

**Q.**

A body of mass 10 kg is dropped to the ground from a height of 10 metres. The work done by the gravitational force is g=9.8 m/sec2

- 490

*Joules*+ 980

*Joules*+ 490

*Joules*- 980

*Joules*

**Q.**In a particular system of unit, if unit of mass becomes half and that of time becomes twice, then 10 joule of work will be written as --- units of work.

**Q.**Two springs A and B are identical except that A is stiffer than B i.e., kA>kB. If the two springs are stretched by the same force, then

- More work is done on B, i.e., WB>WA
- More work is done on A, i.e., WA>WB
- Work done on A and B are equal
- Work done depends on the way in which they are stretched

**Q.**The natural length of spring is 0.3 m and its spring constant is 30 N/m. Suppose we apply an external force F to stretch the spring. How much work is done by the external force as the spring is stretched from 0.1 m to 0.2 m?

- 0.15 J
- 0.3 J
- 0.45 J
- None

**Q.**

In the figure ball A is released from rest, when the spring is at its natural length. For the block B of mass M to leave contact with the ground at some stage, the minimum mass of A must be

2 M

M

A function of M and force constant of spring

M2

**Q.**A 15 gm ball is shot from a spring gun whose spring has a force constant of 600 N/m. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression is (g=10m/s2)

- 6.0 m
- 12.0 m
- 10.0 m
- 8.0 m

**Q.**

The force constant of a wire is k and that of another wire is 2k. When both the wires are stretched through same distance, then the work done

**Q.**

A body of mass 10 kg is dropped to the ground from a height of 10 metres. The work done by the gravitational force is g=9.8 m/sec2

- 490 Joules

+ 490 Joules

- 980 Joules

+ 980 Joules

**Q.**

A spring when stretched by 2 mm its potential energy becomes 4 J. If it is stretched by 10 mm, its potential energy is equal to

54

*J*415

*J*None

4

*J*

**Q.**Natural length of a massless spring (of spring constant k) is x. It is slowly stretched by applying an external force. What is the work done in slowly increasing its extension from 3x to 4x?

- 2.5kx2
- 1.5kx2
- −3.5kx2
- 3.5kx2

**Q.**When 100J of work is done on a fly wheel, it's angular velocity is increased from 60rpm to 180rpm. What is theomens of inertia

**Q.**A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a constant Force 'F' and if maximum displacement of block from its initial position of rest is δ then

- Work done by force F is equal to
- Increases in energy stored in spring is

**Q.**A spring force constant k is cut in two parts at its one third length. when both the parts are streatched by same amount. The work done is

- equal in both
- greater for the longer part
- greater for the shorter part
- data insufficient

**Q.**In the given figure, a movable wire PQ starts moving on the fixed rail at t = 0 with constant speed of 32 m/s. Then work done by external force on wire PQ in 2 s is

**Q.**

Two springs have their force constant as k1 and k2(k1>k2) . When they are stretched by the same force

More work is done in case of second spring

Equal work is done in case of both the springs

No work is done in case of both the springs

More work is done in case of first spring

**Q.**

A $40mm$ long spring is stretched by applying a force. If $10N$ force is required to stretch the spring through one mm, then work done in stretching the spring through $40mm$ is:

**Q.**When two mutually perpendicular simple harmonic motions of same frequency , amplitude and phase are superimposed

- the resulting motion is uniform circular motion.
- the resulting motion is a linear simple harmonic motion along a straight line inclined equally to the straight lines of motion of component ones.
- the resulting motion is an elliptical motion, symmetrical about the lines of motion of the components.
- the two S.H.M. will cancel each other.

**Q.**A wire fixed at the upper end is stretched by 1 m by applying a force F. The work done in stretching is

- F
- 2F
- F2
- F2

**Q.**

Two springs of spring constants $1500N{m}^{-1}$ and $3000N{m}^{-1}$ respectively are stretched with the same force. They will have potential energy in the ratio:

$1:2$

$2:1$

$1:4$

$4:1$

**Q.**

What type of work is done in stretching a Spring?

**Q.**When a 1.0kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g = 10m/s2

- 1.5 Joule
- 2.0 Joule
- 2.5 Joule
- 3.0 Joule

**Q.**

The potential energy of the spring when stretched ______.

**Q.**

If the force constant of a wire is *K*, the work done in increasing the length of the wire by *l* is

*Kl*/2*Kl*

**Q.**

A man raises a mass $m$ to a height $h$ and then shifts it horizontally by a length$x$. What is the work done against the force of gravity?

**Q.**In an experiment it was found that string vibrates in n loops when a mass M is placed on the pan. The mass that should be placed on the pan to make it vibrate in 2n loops with same frequency (neglect the mass of pan) is

- 2M
- M4
- 4M
- M2

**Q.**A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

- mg(h+d)−12kd2
- mg(h−d)−12kd2
- mg(h+d)+12kd2
- mg(h−d)+12kd2