Question

A position-dependent force $F=4x^3+3x^2+2x-4$ newton acts on a small body of mass $2kg$ and displaces it from $x=0$ to $x=2m$. Calculate the work done.

• A. $20J$
• B. $24J$
• C. $16J$
• D. $12J$

Answer 1

Answer: (A)

$20J$

$Workdone={\int }_{x_1}^{x_2}Fdx$

$={\int }_0^{2}4x^3+3x^2+2x-4dx$

$={[\frac{4x^4}{4}+\frac{3x^3}{3}+\frac{2x^2}{2}-4x]}_0^2$

$=[x^4+x^3+x^2-4x{]}_0^2$

$=2^4+2^3+2^2-4\left(2\right)$

$=16+8+4-8$

$=20J$