Question

A 58.4-kg person, running horizontally with a velocity of +3.97 m/s, jumps onto a 14.8-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:0.017

Explanation:

Given

mass of person $m_1=58.4 kg$

mass of sled $m_2=14.8 kg$

velocity of person $v_1=3.97 m/s$

let u be the combined velocity of person and sled

conserving momentum

$m_1v_1=(m_1+m_2)u$

$u=\frac{m_1}{m_1+m_2}\times v_1$

$u=\frac{58.4}{58.4+14.8}\times 3.97$

$u=3.16 m/s$

(b)Total distance traveled by sled and man is 30 m

deceleration Provided by surface is $\mu _k\cdot g$

where $\mu _k$ is coefficient of kinetic friction

using $v^2-u^2=2as$

$0-3.16^2=2(-\mu _k\cdot 9.8)\cdot 30$

$\mu _k=0.0169$