Project 5 Solution: Part B: Escape from Korath

First, it is a good idea to determine the length of the ramp: $$L=\frac{h}{\sin\theta}\approx 6.58\,{\rm m}.$$

Now, we need to determine the kinetic coefficients of friction on the ramp $\mu_{{\rm k}, L}$ and on the ground $\mu_{{\rm k}, D}$. To do this, the students need to drop the second cart down the ramp from rest. After doing this, tell them that it takes $t_{L}=2.42\,{\rm s}$ to drop to the bottom of the ramp, and an additional time $t_{D}=6.91\,{\rm s}$ to come to a halt on the ground.

Using the constant acceleration kinematic equation $x = \frac{1}{2}at^{2}$ down the ramp with the acceleration coming from \begin{eqnarray*}F_{\rm net,ramp} &=& F_{{g},{\rm ramp}}-F_{\rm fric,ramp} \\ ma_{\rm net,ramp}&=&mg\sin\theta-mg\mu_{{\rm k},L}\cos\theta \\ a_{\rm net,ramp}&=&g(\sin\theta-\mu_{{\rm k},L}\cos\theta)\end{eqnarray*} down the ramp, we can solve $$L=\frac{1}{2}g(\sin\theta-\mu_{{\rm k}, L}\cos\theta)t_{L}^{2}$$ for the necessary coefficient: $$\mu_{{\rm k}, L}=\tan\theta-\frac{2L\sec\theta}{g t_{L}^{2}}\approx 0.12.$$

Now we consider the constant acceleration kinematic equation $v_{f}=v_{i}+at$ along the ramp and the ground. To find $\mu_{{\rm k}, D}$ we substitute $\mu_{{\rm k}, L}$ into $$v_{m}=g(\sin\theta-\mu_{{\rm k}, L}\cos\theta)t_{L},$$ and plug this into $$0=v_{m}-\mu_{{\rm k}, D}gt_{D},$$ where $v_{m}$ is the speed at the point where the ramp touches the ground, the cart initially starts from rest at the top of the ramp, comes to a stop on the flat ground, and the acceleration along the flat ground comes from \begin{eqnarray*}F_{\rm net,ground} &=& F_{\rm fric,ground} \\ ma_{\rm net,ground}&=&mg\mu_{{\rm k},D}\\ a_{\rm net,ground}&=&-g\mu_{{\rm k},D}.\end{eqnarray*} Solving for the desired (eliminating $v_{m}$ from the two equations): $$\mu_{{\rm k}, D}=\frac{2L}{gt_{L}t_{D}}\approx 0.08.$$

Now that we have our coefficients, we can set up our condition on the initial velocity. We combine $v_{f}^{2}=v_{i}^{2}+2ad$ twice, piecing together the ramp and the ground. That is, we use $$v_{m}^{2}=v_{i}^{2}+2a_{\rm net, ramp}L\quad\mbox{and}\quad v_{f}^{2}=v_{m}^{2}+2a_{\rm net, ground}D$$ eliminating the velocity at the point where the ramp touches the ground $v_{m}$ ($v_{i}$ is the initial speed at the top of the ramp and $v_{f}$ is the speed at the portal gate): \begin{eqnarray*} v_{i}^{\rm min}&=&\sqrt{(v_{f}^{2})^{\rm min}-2a_{\rm net,ramp}L-2a_{\rm net,ground}D} \\ v_{i}^{\rm min}&=&\sqrt{(v_{f}^{\rm min})^{2}-2g\big((\sin\theta-\mu_{{\rm k}, L}\cos\theta)L-\mu_{{\rm k}, D}D\big)}\approx 12.67\,{\rm m/s}.\end{eqnarray*} Notice that the accelerations coming from the friction are negative (they push against it) and the acceleration coming from gravity is positive (it pulls it down the ramp). Given that the acceleration down the ramp is positive, we can see that $v_{i}<v_{m}$ and $v_{f}<v_{m}$. Furthermore, given the necessary final speed, we see that $v_{f}<v_{i}$.

Tutor Questions:

• Question: Does the frictional force depend on the mass of the object? What about the contact area? Why?
• Expected Answer: Yes mass, no area. The model we have for friction force compresses the materials more when there is more mass by a single object of the same contact area. The fact that there's no effect of area is actually challenging to explain because by keeping the mass the same, but increasing the area means you compress the materials less but over a greater area - the effect is not net change in the frictional force. Friction is actually a much more fascinating subject that is studied in detail (i.e., Tribology)

• Question: Is this problem mass dependent?
• Expected Answer: No.

• Tutor Question: Is your calculated initial velocity reasonably attainable? How do you know?
• Expected Answer: $12\,{\rm m/s}\approx 30\,{\rm mph}$, so no. Let them look up something like how fast olympic sprinters can run.

• Tutor Question: How does the kinetic coefficient differ from the static coefficient of friction? Why are they different, what is happening microscopically?
• Expected Answer: The static coefficient $>$ kinetic coefficient. Students should refer back to the microscopic “roughness” of surfaces and how the asperities interact.

• Tutor Question: Why are anti-lock breaks useful?
• Expected Answer: They increase the amount of time that the static coefficient comes into play.

• Tutor Question: Can you draw free body diagrams from the cart on the ramp and on the ground?
• Expected Answer: …

Qr code for Escape from Korath 2