A body of 20 kg is lying at rest. Under the action of a constant force, it gains a speed of 7 m/s. The work done by the force will be
A
:
490J
B
:
500J
C
:
390J
D
:
430J
General Science
Physics
Motion
Answer
Option 1 : 490J
Explanation
The correct answer is 490J
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) = 20 kg
Final Velocity (v) = 7 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 × 20 × 72
⇒ W = 10 × 49
⇒ W = 490J