Question

a 0.15 kg block initially moving at 3.0 m/s at the base of a ramp set at a 24 degree. if the coefficient of kinetic friction, mu_k = 0.17, how far up the ramp will the block rise before coming to rest? assume the acceleration due to gravity g = 10 m/s^2.

Explanation:

Step1: Analyze forces

The forces along the ramp are the component of gravitational force \(mg\sin\theta\) (down - ramp) and the frictional force \(f = \mu_{k}N\). The normal force \(N=mg\cos\theta\). So the net force \(F_{net}=mg\sin\theta+\mu_{k}mg\cos\theta\). Using \(F = ma\), we get \(a = g\sin\theta+\mu_{k}g\cos\theta\).
Substitute \(\theta = 24^{\circ}\), \(\mu_{k}=0.17\), \(g = 10m/s^{2}\):

$$
LATEXBLOCK0
$$

Step2: Use kinematic equation

We use the kinematic equation \(v^{2}=v_{0}^{2}-2ax\) (negative because it's decelerating). At rest \(v = 0\), \(v_{0}=3.0m/s\).

$$
LATEXBLOCK1
$$

Answer:

\(0.80\), \(m\)