A force of $\left(6 x^2-4 x+3\right) \mathrm{N}$ acts on a body of mass 0.75 kg and displaces it from $x=2…
A force of $\left(6 x^2-4 x+3\right) \mathrm{N}$ acts on a body of mass 0.75 kg and displaces it from $x=2 \mathrm{~m}$ to $x=5 \mathrm{~m}$. The work done by the force is
201 J
215 J
229 J
307 J
Solution
$F=\left(6 x^2-4 x+3\right) N, m=0.75 \mathrm{~kg}, x_1=2 m, x_2=5 \mathrm{~m}$
Work done, $W=\int_{\mathrm{x}_1}^{\mathrm{x}_2} F d x=\int_2^5\left(6 \mathrm{x}^2-4 \mathrm{x}+3\right) \mathrm{dx}$
$=\left[6 \cdot \frac{x^3}{3}-4 \frac{x^2}{2}+3 x\right]_2^5=201 J$