First slide

Work Energy Theorem

Question

A body of mass 2 kg moves down the quadrant of a circle of radius 4 m. The velocity on reaching the lowest point is $8 m s^{- 1}$ . Work done against friction is (take g = $10 m / s^{2}$ )

Moderate

A

78.4 J
B

64 J
C

16 J
D

zero

Solution

According to work energy theorem, work done by all forces $= Δ K E$
$\begin{array}{l} W_{g r} + W_{f r} = Δ K . E = \frac{1}{2} m v^{2} - 0 = \frac{1}{2} \times 2 \times 64 = 64 \\ \Rightarrow W_{f r} = 64 - W_{g r} = 64 - m g R = 64 - 80 = - 16 J \end{array}$
Thus work done by body against friction = 16 J

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