Question

A single conservative force F(x) = b x + a acts on a 4.37 kg particle, where x is in meters, b = 6.09 N/m and a = 5.96 N. As the particle moves along the x axis from x1 = 0.652 m to x2 = 5.1 m, calculate the work done by this force. Answer in units of J.

Answer 1

Answer:104.41 J

Explanation:

Given

Force acting $F(x)=bx+a$

mass of particle $m=4.37 kg$

$a=5.96 N$

$b=6.09 N/m$

Work done is given by

$W=\int_{a}^{b}F.dx$

$W=\int_{0.652}^{5.1}\left ( 6.09x+5.96\right )dx$

$W=\left [ 6.09\times \left ( \frac{x^2}{2}\right )+5.96\times x\right ]_{0.652}^{5.1}$

$W=77.906+26.51 J$

$W=104.41 J$