Question

A $70 \, kg$ man standing on ice throws a $3 \, kg$ body horizontally at $8 \, m \, s^{- 1}$ . The friction coefficient between the ice and his feet is $0.02$ . The distance, through which the man slip is

• A. $0.3\,m$
• B. $2\,m$
• C. $1\,m$
• D. $\infty$

Answer 1

Answer: (A)

At the time of throwing body, velocity gained by man is
$v_{0}=\frac{3 \times \, 8}{70}=\frac{24}{70}$ m/s
$\therefore $ Stopping distance $s=\frac{v_{0}^{2}}{2 \mu g}$
Or $s=\frac{2 4 \times 24}{70 \times 70 \times 2 \times 0.02 \times 10}$
$=\frac{576}{49 \times 40}=\frac{144}{490}=0.3 \, m$