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

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) ∞

Tardigrade Admin · Accepted Answer

At the time of throwing body, velocity gained by man is v0=(3 × 8/70)=(24/70) m/s ∴ Stopping distance s=(v02/2 μ g) Or s=(2 4 × 24/70 × 70 × 2 × 0.02 × 10) =(576/49 × 40)=(144/490)=0.3 m

Q. A $70 k g$ man standing on ice throws a $3 k g$ 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

$0.3 m$

$2 m$

$1 m$

$\infty$

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

Q. A 70kg man standing on ice throws a 3kg body horizontally at 8ms−1 . The friction coefficient between the ice and his feet is 0.02 . The distance, through which the man slip is

0.3m

2m

1m

∞