Question

A 70 kg man standing on ice throws a 3 k body horizontally at 8 m/s. The friction coefficients between the ice and his feet are 0.02. The distance, the man slips is:
A. 0.3 m
B. 2 m
C. 1 m
D. $\infty$

Answer 1

Determine the acceleration of the man produced due to frictional force. Calculate the velocity of the man using law of conservation linear momentum. Use a kinematic equation to determine the distance, the man slips.

Formula used:
Frictional force, $f = \mu Mg$
Here,$\mu$ is the coefficient of friction, M is the mass and g is the acceleration due to gravity.
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.

Complete step by step solution:
We have given that the mass of the man is $M = 70\,kg$, the mass of the body is $m = 3\,kg$ and the coefficient of friction is $\mu = 0.02$.
When the man throws a body in the forward direction, he slips in the opposite direction. The slipping of the man is due to friction force between his feet and surface of ice. We can calculate the magnitude of this force as follows,
$f = \mu Mg$
Here, g is the acceleration due to gravity.
Substituting $\mu = 0.02$, $M = 70\,kg$ and $g = 10\,m/{s^2}$ in the above equation, we get,
$f = \left( {0.02} \right)\left( {70} \right)\left( {10} \right)$
$\Rightarrow f = 14\,N$
We can use Newton’s second law to determine the acceleration of the man as follows,
${F_{net}} = Ma$
$\Rightarrow f = Ma$
$\Rightarrow a = \dfrac{f}{M}$
Substituting $f = 14\,N$ and $M = 70\,kg$ in the above equation, we get,
$a = \dfrac{{14}}{{70}}$
$\Rightarrow a = 0.2\,m/{s^2}$
Now, we can use conservation of linear momentum to determine the final velocity of the man as follows,
$M{u_M} + m{u_b} = M{v_M} + m{v_b}$
Here, u and v represents initial and final velocities and the subscripts “M” and “b” represents velocities of the man and the body.
Since the initial momentum of the system was zero, we can write the above equation as follows,
$0 = M{v_M} + m{v_b}$
$\Rightarrow M{v_M} = - m{v_b}$
$\Rightarrow {v_M} = \dfrac{{m{v_b}}}{M}$
We can ignore the negative sign to determine the magnitude of the velocity.
Substituting $m = 3\,kg$, ${v_b} = 8\,m/s$ and $M = 70\,kg$ in the above equation, we get,
${v_M} = \dfrac{{\left( 3 \right)\left( 8 \right)}}{{70}}$
$\Rightarrow {v_M} = 0.34\,m/s$
This is the initial velocity of the man for his motion in the backward direction.
Now, we can use kinematic equation to determine the distance travelled by the man as follows,
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.
Since $v = 0$, the above equation becomes,
${u^2} = 2as$
$\Rightarrow s = \dfrac{{{u^2}}}{{2a}}$
Substituting $u = 0.34\,m/s$ and $a = 0.2\,m/{s^2}$ in the above equation, we get,
$s = \dfrac{{{{\left( {0.34} \right)}^2}}}{{2\left( {0.2} \right)}}$
$\Rightarrow s = 0.289$
$\Rightarrow s \approx 0.3\,m$
Therefore, the distance the man slips is 0.3 m.

So, the correct answer is “Option A”.

Note:
We have ignored the negative sign each time it is encountered in the solution. The negative sign for the velocity represents the velocity of the man is opposite to the velocity of the body. The distance travelled by the man is also negative. The final velocity of the man will be zero after the friction force halts its motion.