Physics - W.E.P. Question with Solution | TestHub

The net work done on the object is equal to the change in the kinetic energy of the object

$W_{net}= K_f-K_i=\Delta K$

Kinetic energy$K=\frac{1}{2}m{v}^2$

Velocity$v^2=v_x^2+v_y^2$

$=\left(36+4\right){\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$

$v^2=40{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$

$K_f=\frac{1}{2}\times 3\times 40=60 J$

Again, velocity$v=\sqrt{8^2+4^2}$

$v=\sqrt{64+16}=\sqrt{80}\text{ m/s}$

Kinetic energy$K_f=\left(\frac{1}{2}\right)\left(3\right)\times 80{\text{ m}}^2{\text{/s}}^2=120 \mathrm{J}$

From work energy theorem

$W_{net}= K_f-K_i$

$=120 \mathrm{J}-60 \mathrm{J}=60 \mathrm{J}$