Question

How much work is needed to be done to increase the velocity of an object of mass 2 kg from 3 m/s to 6 m/s?

Answer 1

Mass of object = 2kg (m)

Initial velocity = 3m/s (u)

Final velocity = 6m/s (v)

By Work Energy Theorem

$\large\green{\boxed{W= \triangle KE}}$

Where

• W is work done
• ∆KE is change in Kinetic Energy.

$\implies W = \triangle KE$

$\implies W = \dfrac{1}{2}mv^{2}-\dfrac{1}{2}mu^{2}$

$\implies W=\dfrac{1}{2}m(v^{2}-u^{2})$

$\implies W = \dfrac{1}{2}\times2[(6m/s) ^{2}-(3m/s) ^{2}]$

$\implies W = \dfrac{1}{\cancel{2}} ×\cancel{2}(36m^{2} s^{-2}-9m^{2}s^{-2})$

${\underline{\boxed{.°. W = 25J}}}$

Therefore the work done is 25J.