Question

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

• A. 1.4
• B. 4.1
• C. 0.14
• D. 0.41

Answer 1

Answer: (A)

1.4

Answer 2

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