Question

A mass of $2\,kg$ slides down a quarter circular track of radius $1\,m$. If the speed of mass at the bottom is $4\,ms^{-1}$, work done by frictional force is

• A. - 8 J
• B. - 4 J
• C. - 6 J
• D. - 32 J

Answer 1

Answer: (B)

- 4 J

Answer 2

Velocity at lowest point, $\upsilon = \sqrt{2gR}$ K.E.= $\frac{1}{2}m \upsilon^2 = \frac{1}{2} m2gR = mgR = 2 \times 10 \times 1 = 20 \, J$ Real K.E. =$\frac{1}{2} \times 2 \times 4^2$ = 10 J frictional work = 16 - 20 = - 4 J