Question

constant force is applied on a body of 2 kg to give it a displacement s = t^2/2. Work done by agent applying the force upto time t=3 sec is A. 3 J C. 18 J B. 9J
answer with explanation plz​

Answer 1

Explanation:

I think answer is very obvious ok

Answer 2

Given : constant force is applied on a body of mass 2 kg.

             Displacement S = t^2/2.

To Find : Work done by agent applying the force up to time t=3 sec

Solution :

By using work energy theorem,

$\[W = \Delta KE = \frac{1}{2}m{v_2}^2 - \frac{1}{2}m{v_1}^2\]$

$\[S = \frac{{{t^2}}}{2}\]$

On differentiating,

$\[\frac{{ds}}{{dt}} = t\]$

v = t

As, when body is at t = 0s , then v = 0 m/s

     when body is at t = 3s, then v = 3 m/s

Then,

Work done ,

$\[W = \Delta KE = \frac{1}{2}m{v_2}^2 - \frac{1}{2}m{v_1}^2\]$

$\[W = \frac{1}{2} \times 2 \times {\left( 3 \right)^2} - \frac{1}{2} \times 2 \times {\left( 0 \right)^2}\]$

W = 9J

Hence, Work done by agent applying the force up to time t=3 sec is 9J.