Question

The heavier block in an Atwood machine has mass twice that of the lighter one. The tension in the string is 16⋅0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Tension}\mathrm{in}\mathrm{the}\mathrm{string},T=16\mathrm{N} \\ \mathrm{From}\mathrm{the}\mathrm{free}-\mathrm{body}\mathrm{diagrams}, \\ T-2\mathit{}mg+2\mathit{}ma=0\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{\left(}i\mathit{\right)} \\ T\mathit{-}mg\mathit{-}ma\mathit{=}\mathit{0}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{\left(}ii\mathit{\right)} \\ \mathrm{From}\text{e}\mathrm{quation}\text{s}\mathit{}\mathit{\left(}i\mathit{\right)}\mathit{}\mathrm{and}\mathit{}\mathit{\left(}ii\mathit{\right)}\mathit{,} \\ T=4\mathit{}ma \\ \Rightarrow a=\frac{T}{4m}=\frac{4}{m}\mathit{}\mathrm{m}\mathit{/}{\mathrm{s}}^{\mathit{2}} \end{gathered}$$





$$\begin{gathered} \mathrm{Now},\mathrm{S}=ut+\frac{1}{2}a{t}^2 \\ =\left(\frac{1}{2}\right)\times \frac{4}{m}\times 1\left[\mathrm{as}u=0\right] \\ =\left(\frac{2}{m}\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Net}\mathrm{mass}=2m\mathit{-}m\mathit{=}m \\ \mathrm{Decrease}\mathrm{in}\mathrm{potential}\mathrm{energy}, \\ \mathrm{P}.\mathrm{E}.=mgh \\ =m\times g\times \left(\frac{2}{m}\right) \\ =9.8\times 2=19.6\mathrm{J} \end{gathered}$$