question_answer

A body of mass 5 kg starting from point 'A' which is it at a height of 5m from the ground, follows the path ABC as shown in the figure and comes to rest on reaching C, which is at a height of 3m from the ground. If the length of the path ABC is 40m, find the resistance force that makes the body stop at C.

Answer:

Given, \[{{h}_{1}}=5m{{h}_{2}}=3m\] \[m=5kgS=40m\] \[F=?\] P.E. at \[A=mg{{h}_{1}}\,\,P.E.\] at \[C=mg{{h}_{2}}\] Loss in \[P.E.=mg{{h}_{1}}-mg{{h}_{2}}\]                 \[=mg({{h}_{1}}-{{h}_{2}})=F\times S\] \[\Rightarrow \,\,5\times 10(5-3)=40\times F\] \[F=\frac{100}{40}2.5\,\,N\]