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Q.

A piece of ice of mass 50 kg is pushed with a velocity of 5 m/sec along a horizontal surface. As a result of friction between the piece and the surface, the piece stops after traveling 25 m; the ice melted in this process is

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a

50 kg

b

50 gm

c

7.86 gm

d

1.86 gm

answer is D.

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Detailed Solution

$\frac{1}{2} {mv}^{2} = {Jm}_{ice} L \Rightarrow m_{ice} =$ mass of ice melted
$\begin{array}{l} m_{ice} = \frac{{mv}^{2}}{2 JL} = \frac{50 \times 5 \times 5}{2 \times 4 .2 \times 10^{3} \times 80} \\ = 1 .86 \times 10^{- 3} kj \\ = 1 .86 gm \end{array}$

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