Question

In the given figure, two blocks of masses $M=2kg$ and $m=1kg$ start from rest when $\theta ={30}^o$. Pulley is ideal and at height of $2m$ from ground. Friction is absent everywhere. [Take $g=10m/s^2$]

• A. Speed of 1 kg block when $\theta ={45}^o$ is 4.8 m/s
• B. Speed of 2 kg block when $\theta ={45}^o$ is 3.4 m/s
• C. Work done on 1 kg mass during ${30}^o⩽\theta ⩽{45}^o$ is 11.7 J
• D. Power of gravitational force at $\theta ={45}^o$ is 68 W.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Speed of 1 kg block when $\theta ={45}^o$ is 4.8 m/s
B Speed of 2 kg block when $\theta ={45}^o$ is 3.4 m/s
C Work done on 1 kg mass during ${30}^o⩽\theta ⩽{45}^o$ is 11.7 J
D Power of gravitational force at $\theta ={45}^o$ is 68 W.

Consider the variables as shown in figure below.$\tan \left({30}^o\right)=\frac{2}{d+x}\tan \left(45^o\right)=\frac{2}{d}d=2andx=2(\sqrt{3}-1)\sin \left({30}^o\right)=\frac{2}{AB}\Rightarrow AB=4m\sin \left({45}^o\right)=\frac{2}{AC}\Rightarrow AC=2\sqrt{2}mMmovesbyy.y=AB-AC=(4-2\sqrt{2})mByWork-EnergyTheorem\Rightarrow 2\times 10(4-2\sqrt{2})=\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 2\times v{\text{'}}^{2}Where,v=\sqrt{2}v\text{'}\left(byconstraintrelation\right)Solvingweget\Rightarrow v^2=23.4orv=4.8m/sandv\text{'}=3.4m/sWorkdoneon1kgmassduring30°⩽\theta ⩽45°=\frac{1\times m\times v^2}{2}=11.7JPower=F.v=(2\times 10\times 3.4)=68W$