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(A) The work done against friction is converted into heat energy, which raises the temperature of the block.Given: Mass $m = 20 \ kg = 20000 \ g$, vel

Vedclass · Accepted Answer

$0.0025$

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(A) The work done against friction is converted into heat energy, which raises the temperature of the block.Given: Mass $m = 20 \ kg = 20000 \ g$, velocity $v = 0.5 \ ms^{-1}$, time $t = 2.1 \ s$, coefficient of friction $\mu = 0.1$, specific heat $c = 0.1 \ cal \ g^{-1} \ ^{\circ}C^{-1} = 0.1 	imes 4.2 \ J \ g^{-1} \ ^{\circ}C^{-1} = 0.42 \ J \ g^{-1} \ ^{\circ}C^{-1}$.Frictional force $f = \mu mg = 0.1 	imes 20 	imes 10 = 20 \ N$.Work done against friction $W = f 	imes d = f 	imes (v 	imes t) = 20 	imes 0.5 	imes 2.1 = 21 \ J$.Heat energy $Q = mc \Delta T$.Since $W = Q$, we have $21 = 20000 	imes 0.42 	imes \Delta T$.$\Delta T = \frac{21}{20000 	imes 0.42} = \frac{21}{8400} = 0.0025^{\circ} C$.

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