Question

a 72.8 - kg person, running horizontally with a velocity of + 2.62 m/s, jumps onto a 12.5 - 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?

Explanation:

Step1: Apply conservation of momentum for part (a)

The law of conservation of momentum states \(m_1v_1 + m_2v_2=(m_1 + m_2)v\). Here, \(m_1 = 72.8\space kg\), \(v_1=+ 2.62\space m/s\), \(m_2 = 12.5\space kg\), \(v_2 = 0\space m/s\).
Substitute the values into the formula: \((72.8\times2.62)+(12.5\times0)=(72.8 + 12.5)v\)

$$v=\frac{72.8\times2.62}{72.8 + 12.5}$$

$$v=\frac{190.736}{85.3}$$

$$v = 2.24\space m/s$$

Step2: Use work - energy theorem for part (b)

The initial kinetic energy \(K=\frac{1}{2}(m_1 + m_2)v^{2}\), and the work done by friction \(W_f=-\mu_k(m_1 + m_2)gd\).
By the work - energy theorem \(K_f-K_i=W_f\). Since \(K_f = 0\) (comes to rest), \(\frac{1}{2}(m_1 + m_2)v^{2}=\mu_k(m_1 + m_2)gd\)
Cancel out \((m_1 + m_2)\) from both sides: \(\mu_k=\frac{v^{2}}{2gd}\)
Substitute \(v = 2.24\space m/s\), \(g = 9.8\space m/s^{2}\), \(d=30.0\space m\)
\(\mu_k=\frac{(2.24)^{2}}{2\times9.8\times30.0}\)

$$=\frac{5.0176}{588}$$

\(\mu_k=0.00853\)

Answer:

(a) \(2.24\space m/s\)
(b) \(0.00853\)