Question

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

Answer 1

Given mass =0.5 kg and velocity v$=a{x}^{\frac{3}{2}}anda=5m^{\frac{1}{2}}{s}^{-1}$

At x=0 initial velocity u= 0 m/s and x=2 final velocity v$=a{x}^{\frac{3}{2}}=5\times 2^{1.5}=14.14m/s$

Now work done = Change in kinetic energy

$=\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.5\times {14.14}^2=49.98$ Joule