This set of Engineering Physics Assessment Questions and Answers focuses on “Work Done”.

1. A body can have energy without momentum.

a) True

b) False

View Answer

Explanation: A body can have energy without momentum. There is internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum if the momenta of the moving particle may be zero.

2. A particle is acted upon by a force of a constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that ___________

a) Its velocity is constant and kinetic energy is constant

b) Its acceleration and velocity is constant

c) Its kinetic energy is constant and it moves in a circular path

d) Its acceleration is constant and it moves in a circular path

View Answer

Explanation: For a particle in uniform circular motion, the direction of both velocity and acceleration vectors change continuously but their magnitudes remain unchanged. The kinetic energy is not affected. Hence its kinetic energy is constant and it moves in a circular path.

3. An athlete in the Olympic game covers a distance of 100m in 10s. His kinetic energy can be estimated in the range?

a) 200J – 500J

b) 2×10^{5}J – 3×10^{5}J

c) 20000J – 50000J

d) 2000J – 5000J

View Answer

Explanation: Average speed of the athlete, v = s/t = 10m/s

Assuming the mass of the athlete to 60kg, his average kinetic energy would be

K = 1/2×60×10

^{2}

K = 3000J.

4. A particle of mass 100g is thrown vertically upwards with a speed of 5m/s. The work done by the force of gravity during the time the particle goes up is ___________

a) 1.25J

b) 0.5J

c) -0.5J

d) -1.25J

View Answer

Explanation: Work done by the force of gravity = Loss in kinetic energy of the body

Work done by the force of gravity = 1/2 m(v

^{2}-u

^{2}) = 1/2×100/1000×(0

^{2}-5

^{2})J

Work done by the force of gravity = -1.25J.

5. Statement 1: Linear momentum of a system of particles is zero.

Statement 2: Kinetic energy of a system of particles is zero.

a) 1 does not imply 2 and 2 does not imply 1

b) 1 implies 2 but 2 does not imply 1

c) 1 does not imply 2 but 2 implies 1

d) 1 implies 2 and 2 implies 1

View Answer

Explanation: When the linear momentum of a system of particles is zero, the velocities of the individual particles may not be zero. The kinetic energy of the system of particles may be non-zero. Thus 1 does not imply 2. When the kinetic energy of the system of particles is zero, then the kinetic energy and hence the velocity of each particle is zero. Therefore the linear momentum of a system of particles is zero. Thus 2 imply 1.

6. A spherical ball of mass 20kg is stationary at the top of a hill of height 100m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30m and finally rolls done to a horizontal base at a height of 20m above the ground. The velocity attained by the ball is ___________

a) 10m/s

b) 10√30m/s

c) 40m/s

d) 20m/

View Answer

Explanation: Total energy at 100m height = Total energy at 20m height

mgh

_{1}= mgh

_{2}+ 1/2 mv

^{2}

v = √(2g(h

_{1}-h

_{2})) = √(2×10×(100-20)) = 40m/s.

7. If mass-energy equivalence is taken into account when water is cooled to form ice, the mass of water should ___________

a) Increase

b) Remain unchanged

c) Decrease

d) First increase and then decrease

View Answer

Explanation: The heat energy possessed by water gets converted into mass when ice is formed. This increases the mass. Therefore when water is cooled to form ice, then mass of the water should increase.

8. A body moves a distance of 10m under the action of force F = 10N. If the work done is 25 J, the angle which the force makes with the direction of motion is?

a) 0°

b) 30°

c) 60°

d) None of the mentioned

View Answer

Explanation: W = Fscosθ

25 = 10×10cosθ

Cosθ = 1/4

θ = cos

^{(-1)}1/4.

9. Two bodies of mass and 4m have equal kinetic energy. What is the ratio of their momentum?

a) 1:4

b) 1:2

c) 1:1

d) 2:1

View Answer

Explanation: p = √2mK

For same K, p

_{1}/p

_{2}= √(m

_{1}/m

_{2}) = √(m/4m) = 1/2 = 1:2.

**Sanfoundry Global Education & Learning Series – Engineering Physics.**

To practice all areas of Engineering Physics Assessment Questions, __here is complete set of 1000+ Multiple Choice Questions and Answers__.