Question

A 62.2-kg person, running horizontally with a velocity of 3.80 m/s, jumps onto a 19.7-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow

Answer 1

Answer:

a) $v = 2.886\,\frac{m}{s}$, b) $\mu_{k} = 0.014$

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

$(62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v$

$v = 2.886\,\frac{m}{s}$

b) The deceleration experimented by the system person-sled is:

$a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}$

$a = -0.139\,\frac{m}{s^{2}}$

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

$-\mu_{k}\cdot m\cdot g = m\cdot a$

$\mu_{k} = -\frac{a}{g}$

$\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }$

$\mu_{k} = 0.014$