Discussion Forum

A time-dependent force F =6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 s will be 

Answer:

Explanation:

From Newton's second law, $\frac{\triangle p}{\triangle t}=F$

$\Rightarrow \triangle p=F\triangle t$

 $\therefore$          $p=\int_{}^{}dp =\int_{0}^{1}  F dt$

$\Rightarrow  p= \int_{0}^{1} 6t dt=3kg(\frac{m}{s})$

Also , change in kinetic energy 

 $\triangle k= \frac{\triangle p^{2}}{2m}=\frac{3^{2}}{2\times 1}$

 = 4.5

From work -energy theorm, work done = change in kinectic energy

    So, work done = Δ k= 4.5 J