Question

A wooden box of 6 kg is lying at rest. Under the action of a constant force, it gains a speed of 8 m/s. The work done by the force will be ________.
1. 195 J
2. 192 J
3. 200 J
4. 180 J

Answer 1

Correct Answer - Option 2 : 192 J

The correct answer is 192 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
• \(W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = {{Δ }}K\)
  • Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

Given,

Mass (m) = 6 kg

Final Velocity (v) = 8 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 × 6 × 8²

⇒  W = 192 J