Question

A force F is related to the position of a particle by the relation $F=\left(10x^2\right)N$. The work done by the force when the particle moves from x=2m to x=4 m is

• A. $\frac{56}{3}J$
• B. $560J$
• C. $\frac{560}{3}J$
• D. $\frac{3}{560}J$

Answer 1

Answer: (C)

$\frac{560}{3}J$

$F=10x^{2}N$

We know that work done $={\int }_{x_1}^{x_2}F.dx$

$\omega ={\int }_2^{4}10x^{2}dx$

$\Rightarrow \omega =10{\int }_2^4{x}^{2}dx$

$=10{\left[\frac{x^3}{3}\right]}_2^4$

$=10[\frac{64}{3}-\frac{8}{3}]=\frac{560}{3}J$