Question

A body of mass $10$ kg is under a constant force which causes a displacement S is in metres given the relation $s=\frac{2}{3}{t}^4$, where $+$ is in seconds. To find work-done by the force $5s$.

Answer 1

Given mass=10 kg; and displacement s $=\frac{2}{3}{t}^4$

and force = Fs

Now differentiating with respect to time we get $\frac{ds}{dt}=\frac{8}{3}{t}^{3}dv=\frac{8}{3}{t}^3\frac{dv}{dt}=\frac{8}{3}{t}^3=>a=8t^2$

Now Force F= mass$\times$acceleration=5s

10$\times 8t^2=\frac{10}{3}{t}^{4}t=\sqrt{24}$

Work done W= $=5\left(\frac{2}{3}\right(\sqrt{24}{)}^4)=1920$Joule