A body of 4.0 kg is lying at rest. Under the action of a constant force, it gains a speed of 5 m/s. The work done by the force will be _______.

Question

Question

This question was previously asked in

RRB ALP Previous Paper 2 (Held On: 9 Aug 2018 Shift 2)

Answer 1

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) = 4.0 kg

Final Velocity (v) = 5 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 × 4 × 5²
⇒  W = 2 × 25
⇒  W = 50 J