A machine with efficiency $2 / 3$ used 12 J of energy in lifting 2 kg block through certain height and it is…
- $\sqrt{2} \mathrm{~ms}^{-1}$
- $2 \mathrm{~ms}^{-1}$
- $2 \sqrt{2} \mathrm{~ms}^{-1}$
- $0.2 \mathrm{~ms}^{-1}$
Solution
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)