Question

A body of 5.0 kg is lying at rest. Under the action of a constant force, it gains a speed of 10 m/s. The work done by the force will be _______.

1. 400J
2. 250J
3. 300J
4. 150J

Answer 1

Correct Answer - Option 2 : 250J

The correct answer is 250J.

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,
Mass (m) = 5.0 kg

Final Velocity (v) = 10 m/s and initial velocity (u) = 0 m/s
According to the work-energy theorem,

⇒ Work done = Change in K.E
⇒ W = Δ K.E

Since initial speed is zero so the initial KE will also be zero.

⇒ Work done (W) = Final K.E = 1/2 mv²
⇒ W = 1/2 × 5 × 10²
⇒ W = 5× 50
⇒ W = 250 J