Question

A force of $F=(5y+20)\hat{j}\text{N}$ acts on a particle. The work done by this force when the particle is moved from $y=0\text{m}\text{to}y=10\text{m}$ is ______$\text{J}$.

Answer 1

Given that,

$F=(5y+20)\hat{j}$

As the force is varying with position, Work done is given by,

$W=\int Fdy={\int }_0^{10}(5y+20)dy$

$\Rightarrow W={[\frac{5y^2}{2}+20y]}_0^{10}$

$\Rightarrow W=\frac{5}{2}\times 100+20\times 10$

$\Rightarrow W=250+200=450\text{J}$

Hence, $450$ is the correct answer.