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A block of mass 2 kg is released from A on the track that is one quadrant of a circle of radius 1 m as shown in
fig. It slides down the track and reaches B with

speed of4 m/s and finally stops at C at a distance of 3 m from B The work done against the force of friction is

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-2J

-20J

10J

answer is B.

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Detailed Solution

$W_{f} = ΔU + ΔK = (ΔU)_{A \to B} + (ΔU)_{B \to C} + (ΔK)_{A \to B} + (ΔK)_{B \to C} = (U_{B} - U_{A}) + (U_{C} - U_{B}) + (K_{B} - K_{A}) + (K_{C} - K_{B}) = (U_{C} - U_{A}) + (K_{C} - K_{A}) = 0 - (2 \times 10 \times 1) + (0 - 0) = - 20 J$

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