A block of ice of mass $50 k g$ is pushed out on a horizontal plane with a velocity of $5 m / s$ . Due to friction it comes to rest after covering a distance of $25 m$ . How much ice will melt?

1 g

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2.50 g

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1.87 g

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0.86 g

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Solution

The correct option is C $1.87 g$
Energy with the block of ice $= \frac{1}{2} m v^{2}$
Energy with the block $= \frac{1}{2} \times (50 k g) (5)^{2} = 625 J$
Latent heat of fusion of ice $(L_{f u s i o n})$ = $333.5 J / g$
Melted mass is given by:
$m L_{f u s i o n} = 625 J$
i.e. $m = \frac{625}{L_{f u s i o n}} = 1.87 g$

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A block of ice of mass 50 kg is pushed out on a horizontal plane with a velocity of 5 m/s. Due to friction it comes to rest after covering a distance of 25 m. How much ice will melt?

The correct option is C 1.87 gEnergy with the block of ice =12mv2Energy with the block =12×(50kg)(5)2=625JLatent heat of fusion of ice (Lfusion) = 333.5J/gMelted mass is given by:mLfusion=625Ji.e. m=625Lfusion=1.87g

A block of ice of mass $50 k g$ is pushed out on a horizontal plane with a velocity of $5 m / s$ . Due to friction it comes to rest after covering a distance of $25 m$ . How much ice will melt?