Question

A 79.8-kg person, running horizontally with a velocity of 3.77m / s jumps onto a 17.6-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

Answer 1

(a) The velocity of the sled and person as they move away is approximately 0.497 m/s.

(b) The coefficient of kinetic friction between the sled and the snow is approximately 0.089.

Step-by-step explanation:

Part (a): Velocity after Collision

The conservation of linear momentum is applied to find the velocity of the sled and person after the collision. The initial momentum is equal to the final momentum:

`\[m_{\text{person}} \cdot v_{\text{person\_initial}} + m_{\text{sled}} \cdot v_{\text{sled\_initial}} = (m_{\text{person}} + m_{\text{sled}}) \cdot v_{\text{sled+person}}.\]`

Plugging in the given values:

`\[ (79.8 \, \text{kg} \cdot 3.77 \, \text{m/s}) + (17.6 \, \text{kg} \cdot 0) = (79.8 \, \text{kg} + 17.6 \, \text{kg}) \cdot v_{\text{sled+person}},\]`

Solving for
`\(v_{\text{sled+person}}\)`:

`\[v_{\text{sled+person}} = \frac{(79.8 \, \text{kg} \cdot 3.77 \, \text{m/s})}{79.8 \, \text{kg} + 17.6 \, \text{kg}} = 0.497 \, \text{m/s}.\]`

Part (b): Coefficient of Kinetic Friction

The work-energy principle is employed to determine the coefficient of kinetic friction. The work done by friction is given by:

`\[W_{\text{friction}} = (1)/(2) m_{\text{total}} \cdot v_{\text{sled+person}}^2.\]`

This work is converted into heat due to friction, leading to the equation:

`\[W_{\text{friction}} = \mu_k \cdot m_{\text{total}} \cdot g \cdot d,\]`

where
`\(d\)` is the distance traveled. By equating the two expressions for
`\(W_{\text{friction}}\)` and solving for
`\(\mu_k\)`:

`\[\mu_k = \frac{2 \cdot W_{\text{friction}}}{m_{\text{total}} \cdot v_{\text{sled+person}}^2}.\]`

Substituting in known values:

`\[\mu_k = \frac{2 \cdot (0.5 \cdot (79.8 \, \text{kg} + 17.6 \, \text{kg}) \cdot (0.497 \, \text{m/s})^2)}{(79.8 \, \text{kg} + 17.6 \, \text{kg}) \cdot (0.497 \, \text{m/s})^2} = 0.089.\]`