Question

A body of mass $0.5kg$ travels in a straight line with velocity $v=a{x}^{3/2}$ where $a=5m^{-1}/2s^{-1}$. What is the work done by the net force during its displacement from $x=0$ 𝑡𝑜 $x=2m$?

Answer 1

Step 1 : $\text{Initial velocity of the body}$

Given:

Mass of body $m=0.5kg$,

velocity $\upsilon =a{x}^{3/2}$ where $a=5m^{-1}/2s^{-1}$,

Initial velocity at $x=0$, $v_1=5\times 0=0m/s$

Step 2 : $\text{Final velocity of the body}$

Final velocity at $x=2$, $v_2=5\times 2^{3/2}=5\times 2.82m/s$

Step 3 : $\text{Applying work energy principle}$

As per work energy principle,

Work done = Increase in kinetic energy

$W=\frac{1}{2}\times 0.5\left[\right(5\times 2.82{)}^2-0]J$

$W=50J$

So, the work done due to the net force for the given displacement is $50J$