Question

A body of mass 0.5 kg0.5kg travels in a straight line with velocity v=5 x^{3 / 2} .v=5x 3/2 . The work done by the net force during the displacement from x=0x=0 to x=2 mx=2m is ________(J)(J)

Answer 1

Given :

Mass of the body travelling in a straight line = 0.5 kg

Velocity of the body = $5x^{\frac{3}{2}}$

To Find :

The amount of work done by the net force during the displacement from x = 0 to x = 2 m is = ?

Solution :

∴From Work Energy Theorem , we know that :

Change in kinetic energy = work done

i.e. W = $\Delta KE$

So, W=$(KE)_{x=2} - (KE)_{x=0}$            -(1)

We also know that $KE = \frac{1}{2}mv^2$

So, using this eq 1 becomes :

W$= \frac{1}{2}m(v^2_{2}-v^2_1)$

   

 =  $\frac{1}{2}\times 0.5 \times [(5x^{\frac{3}{2}})^2_{x=2}- (5x^{\frac{3}{2}})^2_{x=0}]$

 $= \frac{1}{2}\times 0.5 \times (5\times 2^{\frac{3}{2} } )^2$

 $= \frac{1}{2}\times 0.5 \times 25 \times 2^3$

  $=\frac{1}{2}\times \frac{5}{10}\times 25 \times 8$

  $= \frac{25 \times 8 }{4}$

 =$25\times 2$

 = 50 J

So , work done by net force during displacement from x = 0 to x = 2 m is 50 J.