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Q. A heavy box having a mass of $300\, kg$ is pulled along the floor for $10\, m$. If the coefficient of sliding friction is $0.2$, kilocalories of heat produced is

Thermal Properties of Matter

Solution:

Here, m = 0.2, $m$ = 300 $kg, g$ = 9.8 $m/s^2$,
s = 10 $m, w$ = ?, Q = ?,
$J =4.2 \times 10^3$ joule/k cal.
Now, work done in moving the box against friction is given by w = F $\times$ s
$\because$ F = $\mu$ R = $\mu \, mg$ $\therefore $ $w = \mu mg \times s = 0.20 \times 300 \times 9.8 \times 10$ = 5880 joule
Again, we know that heat produced Q = $\frac{w}{J}$ $\therefore $ Q = $\frac{5880}{4.2 \times 10^3} $ = 1.4 k calorie