Question 71
CBSE Class 11 Physics · Work, Energy and Power · Work and kinetic energy
A \(2\,\mathrm{kg}\) body starts from rest under a 7 N horizontal force on a table with kinetic-friction coefficient 0.1. Find the work by the applied force, friction and net force in \(10\,\mathrm{s}\); use \(g=9.8\,\mathrm{m\,s}\)⁻².
Solution to Question 71
Complete workingFirst determine friction, acceleration and displacement. Then calculate each work and verify the work–energy theorem.
\[\begin{aligned}f_k &= \mu mg\\&= (0.1)(2)(9.8)\\&= 1.96\ \mathrm N\end{aligned}\]
\[\begin{aligned}a &= \frac{7-1.96}{2}\\&= 2.52\ \mathrm{m\,s^{-2}}\end{aligned}\]
\[\begin{aligned}s &= \frac12at^2\\&= \frac12(2.52)(10)^2\\&= 126\ \mathrm m\end{aligned}\]
\[\begin{aligned}W_F &= 7(126)\\&= 882\ \mathrm J\end{aligned}\]
\[\begin{aligned}W_f &= -1.96(126)\\&= -247\ \mathrm J\end{aligned}\]
\[\begin{aligned}W_{\rm net} &= 882-247\\&= 635\ \mathrm J\\&= \Delta K\end{aligned}\]
Final answer
\[W_F=882\ \mathrm J,\ W_f=-247\ \mathrm J,\ W_{\rm net}=\Delta K=635\ \mathrm J\]