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 _______.

This question was previously asked in
RRB ALP Previous Paper 2 (Held On: 9 Aug 2018 Shift 2)
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  1. 40J
  2. 50J
  3. 30J
  4. 60J

Answer (Detailed Solution Below)

Option 2 : 50J
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Detailed Solution

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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

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

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