Question

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

• A. 1.4
• B. 4.1
• C. 0.14
• D. 0.41

Answer 1

Answer: (A)

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