Question 1

Which of the following statements correctly defines kinetic energy?

Accepted Answer

Kinetic energy is the energy an object possesses by virtue of its motion; KE = ½mv².

Question 2

A 4 kg block is pushed with a force of 16 N at an angle of 60° to the horizontal direction for a distance of 3 m. Calculate the work done by the applied force.

Accepted Answer

Work = F × d × cos(θ) = 16 × 3 × cos(60°) = 16 × 3 × 0.5 = 24 J.

Question 3

A force of 20 N is applied to move an object 5 m in the direction of the force. Calculate the work done.

Accepted Answer

Work = Force × Displacement × cos(θ); W = 20 × 5 × cos(0°) = 100 J.

Question 4

A 2 kg object is lifted vertically through a height of 10 m. How much work is done against gravity? (Take g = 10 m/s²)

Accepted Answer

Work against gravity = mgh = 2 × 10 × 10 = 200 J.

Question 5

An engine does 5000 J of work in 10 seconds. What is its power output?

Accepted Answer

Power = Work / Time = 5000 / 10 = 500 W.

Question 6

According to the work-energy theorem, the net work done on an object equals its change in:

Accepted Answer

The work-energy theorem states W_net = ΔKE, relating net work to change in kinetic energy.

Question 7

A 5 kg object is raised 4 m and then lowered 2 m from its initial position. Net work done by gravity is: (g = 10 m/s²)

Accepted Answer

Net displacement = -2 m (downward); W_gravity = mgh = 5 × 10 × (-2) = -100 J (negative due to downward displacement).

Question 8

If a 10 W bulb operates for 1 hour, how much energy does it consume?

Accepted Answer

Energy = Power × Time = 10 W × 3600 s = 36,000 J.

Question 9

A ball of mass 0.5 kg is thrown vertically upward with initial velocity 20 m/s. Its kinetic energy at the highest point is: (g = 10 m/s²)

Accepted Answer

At the highest point, velocity = 0 m/s; therefore KE = ½mv² = 0 J.

Question 10

A force of 20 N is applied to move an object 5 m in the direction of the force. Calculate the work done.

Accepted Answer

Work = Force × Displacement × cos(θ); W = 20 × 5 × cos(0°) = 100 J.

Question 11

A 2 kg object is lifted vertically upward by 10 m. How much work is done against gravity? (g = 10 m/s²)

Accepted Answer

Work against gravity = mgh = 2 × 10 × 10 = 200 J.

Question 12

A force of 15 N is applied at 60° to the displacement of 8 m. Find the work done.

Accepted Answer

Work = F × d × cos(θ) = 15 × 8 × cos(60°) = 15 × 8 × 0.5 = 60 J.

Question 13

A car of mass 1000 kg accelerates from rest to 20 m/s. What is the kinetic energy gained?

Accepted Answer

KE = ½mv² = ½ × 1000 × (20)² = 200,000 J = 200 kJ.

Question 14

An engine does 5000 J of work in 10 seconds. What is its power output?

Accepted Answer

Power = Work / Time = 5000 / 10 = 500 W.

Question 15

A block of mass 5 kg is pushed across a frictionless horizontal surface by a constant force of 20 N applied at 60° above the horizontal. The block moves 10 m. Calculate the work done by the applied force.

Accepted Answer

Work = F·cos(θ)·s = 20 × cos(60°) × 10 = 20 × 0.5 × 10 = 100 J.

Question 16

A 2 kg object is lifted vertically upward from rest through a height of 5 m. If the lifting force is constant and the object reaches a velocity of 4 m/s at the top, find the work done by the lifting force (g = 10 m/s²).

Accepted Answer

By work-energy theorem: W_net = ΔKE; W_lift - mg·h = ½mv² - 0, so W_lift = 100 + 16 = 116 J.

Question 17

A motor delivers 1500 W of mechanical power to lift a load. How long will it take to lift a 300 kg load to a height of 20 m at constant velocity (g = 10 m/s²)?

Accepted Answer

Work needed = mgh = 300 × 10 × 20 = 60,000 J; Time = Work/Power = 60,000/1500 = 40 s.

Question 18

A 4 kg ball is thrown upward with an initial velocity of 20 m/s. Taking g = 10 m/s², which statement correctly relates work and kinetic energy at maximum height?

Accepted Answer

From ground to maximum height, gravity opposes motion; work by gravity = -mgh = -ΔKE = -(0 - 800) = -800 J, which equals the negative of initial kinetic energy of 800 J.

CDS Work Energy Power

Work Energy Power — Practice Questions