# Work Energy Theorem

## Trending Questions

**Q.**

In physics What is the sign convention used for work done, in thermodynamics.

**Q.**

Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s2. The work done by the (i) gravitational force (ii) resistive force of air is

- (i) 1.25 J, (ii) –8.25 J
- (i) 100 J, (ii) 8.75 J
- (i) 10 J, (ii) –8.75 J
- (i) –10 J, (ii) –8.25 J

**Q.**A pendulum bob has a speed 3 m/s while passing through its lowest position . The length of the pendulum is 0.5 m. The speed of the bob when it makes an angle of 60∘ with the vertical is (Take g=10 m/s2)

- 1 m/s
- 2 m/s
- 3 m/s
- 4 m/s

**Q.**A body of mass 0.5 kg travels in a straight line with velocity, v=ax3/2, where a=5 m−1/2/s. What is the work done by the net force during its displacement from x=0 to x=2 m?

- 30 J
- 40 J
- 50 J
- 20 J

**Q.**A particle of mass m accelerating uniformly has velocity v at time t1. What is work done in time t?

- 12mv2t21t2
- mv2t21t2
- 2mv2t21t2
- 12(mt1)2t2

**Q.**The additional kinetic energy to be provided to a satellite of mass m, revolving around a planet of mass M, to transfer it from a circular orbit of radius 2R to another orbit of radius 3R is:

- GmM12R
- GmM6R
- GmM3R
- GmMR

**Q.**A block is released from rest from a height h=5 m. After travelling through the smooth curved surface, it moves on the rough horizontal surface through a length l=8 m and climbs onto the other smooth curved surface through a height h′. If coefficient of friction of the rough surface is μ=0.5, find h′.

- 5 m
- 1 m
- 2 m
- 4 m

**Q.**A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10, 000 N/m. The spring compresses by

- 2.5 cm
- 5.5 cm
- 11.0 cm
- 8.5 cm

**Q.**

A block of mass m lying on a smooth horizontal surface is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is

- F√mk
- πF√mk
- 2F√mk
- Fπ√mk

**Q.**The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is

**Q.**Power supplied to a particle of mass 2 kg varies with time as P=3t22 W, where t is in seconds. If velocity of particle at t=0 is v=0, the magnitude of velocity of particle at time t=2 s will be

- 1 m/s
- 2 m/s
- 2√2 m/s
- 4 m/s

**Q.**A stone of mass 1 kg tied to a light inextensible string of length L=103 m is whirling in a circular path of radius L, in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension is 4 and if g is taken to be 10 m/s2, the speed of the stone at the highest point of the circle is

- 20 m/s
- 10√3 m/s

- 5√2 m/s
- 10 m/s

**Q.**A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position, where the string is horizontal, (g being acceleration due to gravity) is

- √2(u2−gl)
- √u2−gl
- u−√u2−2gl
- √2gl

**Q.**

If force $F$, work $W$ and velocity $v$ are taken as fundamental quantities. What is the dimensional formula of time?

**Q.**A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

- logex
- x
- ex
- x2

**Q.**Power applied to a particle varies with time as P=(3t2−2t+1) W, where t is in second. Find the change in its kinetic energy as time changes from t=2 s to t=4 s.

- 61 J
- 32 J
- 102 J
- 46 J

**Q.**For the block moving as shown in figure, maximum compression in the spring will be approximately:

[Take g=10 m/s2, √1552−40√3≈38.50 & 10+2√3≈13.5 for calculation. Neglect the effect of corner]

- 2.5 m
- 3 m
- 2.1 m
- 2 m

**Q.**Velocity-time graph for a body of mass 10 kg is shown in figure. Work done on the body in the first two seconds of motion is

- 12000 J
- −12000 J
- −4500 J
- −9300 J

**Q.**An unloaded bus can be stopped by applying brakes on straight road after covering a distance x. Suppose the passengers add 50% of its weight as the load and the braking force remains unchanged, how far will the bus go after the application of the brakes? (Velocity of bus in both case is same)

- Zero
- 1.5x
- 2x
- 2.5x

**Q.**The position-time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t=4 s?

- 2/3 kg m/s
- −3/2 kg m/s
- 3/2 kg m/s
- −2/3 kg m/s

**Q.**A body of mass 10 kg dropped from a height 20 m, acquires a velocity of 10 m/s when it reaches ground. Magnitude of work done by air resistance (in J) is

**Q.**A particle of mass M, starting from rest, undergoes uniform acceleration. If the speed acquired in time T is V, the average power delivered to the particle is

- MV2T
- 12MV2T2
- MV2T2
- 12MV2T

**Q.**Two springs A and B having spring constant KA and KB (KA = 2KB) are stretched by applying force of equal magnitude. If energy stored in spring A is EA then energy stored in B will be

**Q.**

An object of mass $m$ is moving with a constant velocity, $v$. How much work should be done on the object in order to bring the object to rest?

**Q.**The system is released from rest and the block of mass 1 kg is found to have a speed of 0.3 m/s after it has descended through a distance of 1 m. The value of coefficient of kinetic friction is

- 0.24
- 0.36
- 0.12
- 0.48

**Q.**

The momentum of a body increases by 20%. Calculate the percentage increase in its K.E.

**Q.**A stone with weight W is thrown vertically upward into the air from ground level with initial speed v0. If a constant force f due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is

- h=v202g(1+fW)
- h=v202g(1−fW)
- h=v203g(1+fW)
- h=v202g(1+2fW)

**Q.**An uniform cable of mass M and length L is placed on a horizontal surface such that its (Ln) th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

- 2 MgLn2
- MgLn2
- MgL2n2
- nMgL

**Q.**A block of mass 5 kg is moving in x− direction with a constant speed of 20 m/s. It is subjected to a retarding force of F=−0.1x joules/metre during its travel from x=20 m to x=30 m. Its final K.E. will be:

- 975 J
- 1025 J
- 725 J
- 825 J

**Q.**A body at rest may have:

- speed
- momentum
- energy
- velocity