question_answer

A block of mass 3 kg starts from rest and slides down a curved path in the shape of a quarter-circle of radius \[2\text{ }m\]and reaches the bottom of path with a speed \[1\text{ }m/s\]. If ?g' is \[10\text{ }m/{{s}^{2}},\] the amount of work done against friction is

A) \[60\,\,J\]            

B) \[36\,\,J\]

C) \[24\,\,J\]            

D) \[12\,\,J\]

Correct Answer: B

Solution :

Applying Work-Energy theorem, we can write \[\Delta W=\Delta K\] \[\Rightarrow \] \[{{W}_{t}}+{{W}_{g}}={{K}_{t}}={{K}_{i}}\] Where. \[{{W}_{f}}\] = Work done by frictional force \[{{W}_{g}}\] = Work done by gravity \[{{W}_{f}}\] = Final kinetic energy \[{{W}_{i}}\] = Initial kinetic energy \[\Rightarrow \] \[{{W}_{f}}+mg=\frac{1}{2}m{{v}^{2}}-0\] \[\Rightarrow \] \[{{W}_{f}}=\frac{1}{2}m{{v}^{2}}-mgR\] \[=\frac{1}{2}\times 3\times 16-3\times 10\times 2\] \[=24-60=-36\,J\]