Question

An object of 5 kg is lying at rest. Under the action of a constant force, it gains a speed of 3 m/s. The work done by the force will be _________.
1. 20 J
2. 23 J
3. 22.5 J
4. 24.5 J

Answer 1

Correct Answer - Option 3 : 22.5 J

The correct answer is 22.5 J.

• 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:

Given,

Mass (m) = 5 kg

Final Velocity (v) = 3 m/s 

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 × 3²

⇒  W = 45/2

⇒  W = 22.5 J