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 \[P.E.=mg{{h}_{1}}-mg{{h}_{2}}\] \[=mg({{h}_{1}}-{{h}_{2}})=F\times S\] \[\Rightarrow \,\,5\times 10(5-3)=40\times F\] \[F=\frac{100}{40}2.5\,\,N\]
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