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 m and reaches the bottom of path with a speed 1 m/s. If g is\[10\text{ }m/{{s}^{2}},\]the amount of work done against friction is

A)  60J            

B)  36J

C)  24J            

D)  12J

Correct Answer: B

Solution :

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