Question

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

• A. 9 J
• B. 6 J
• C. 0 J
• D. 3 J

Answer 1

The correct option is C

0 J

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

at t = 0; $v_1=-6m/sandatt=6sec,v_2=6m/s$

So, change in kinetic energy

$=W=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=0$