Question

The displacement $x$ of a particle moving in one dimension, under the action of a constant force is related to time $t$ by the equation $t=\sqrt{x}+3$ where x is in metre and t in second. Calculate the work done by the force in the first $6s$ in $Joule$ :

• A. $10$
• B. $0$
  
• C. $5$
  
• D. $2$
  

Answer 1

The correct option is B $0$

Given equation can be rewritten as,
$x=(t-3{)}^2$

$v=\frac{dx}{dt}=2(t-3)$

$v_f=2(6-3)=6$

$v_i=2(0-3)=-6$

So, from work-energy theorem

$W=\Delta KE=\frac{1}{2}m[v_f^2-v_i^2]=\frac{1}{2}m[6^2-{(-6)}^2]=0$

i.e., work done by the force in the first $6s$ is zero.