A machine with efficiency $2 / 3$ used 12 J of energy in lifting 2 kg block through certain height and it is…

A machine with efficiency $2 / 3$ used 12 J of energy in lifting 2 kg block through certain height and it is allowed to fall through the same. The velocity while it reach the ground is
  1. $\sqrt{2} \mathrm{~ms}^{-1}$
  2. $2 \mathrm{~ms}^{-1}$
  3. $2 \sqrt{2} \mathrm{~ms}^{-1}$
  4. $0.2 \mathrm{~ms}^{-1}$

Solution

Potential energy of the machine, $\begin{aligned} & \mathrm{U}=\frac{2}{3} \times 12=8 \mathrm{~J} \\ & \Rightarrow \mathrm{mgh}=8 \Rightarrow \mathrm{~h}=\frac{8}{2 \times 10}=0.4 \mathrm{~m} \end{aligned}$
By conservation of energy, Loss in PE = Gain in K.E $\begin{aligned} & \Rightarrow \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \mathrm{v}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 0.4}=2 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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