# Finding the distance without using time as a variable

A small child with mass $M$ rides down an icy slope of height $h$ and inclination $\theta$ on a toboggan of mass $m$. At the bottom of the slope they continue to slide on flat ground, slowing and coming to a halt due to friction.

The coefficient of friction between the toboggan and the ground is $\mu_k$. Find an expression for the distance the child and toboggan travel before stopping. (Assume the child starts from rest and the friction on the slope is negligible. Use any variable or symbol stated above as necessary.)

$$\Delta x = \ldots$$

I know that $Δx = \dfrac{h}{\sin\theta} + \ldots$

I can't seem to figure out the 2nd part without using time as a variable.

• Whoever voted down this question could at least tell me why, so that I don't repeat whatever I did that was wrong in the future. Commented Apr 4, 2015 at 0:19
• Have you tried the work-energy theorem? Commented Apr 4, 2015 at 0:25
• The problem is that we haven't covered the energy areas yet. So I don't think it will be involved. But I'll give it a try Commented Apr 4, 2015 at 0:35
• I didn't manage to get the correct answer :/ Commented Apr 4, 2015 at 1:03
• Hi, Jake. You were downvoted most probably because you asked a question that is not on specific physics concept; just homework-sort. It is not a homework-solving site. To know how to post such quo here, go to our meta site.
– user36790
Commented Apr 4, 2015 at 1:04

Let the co-ordinate axes be along the slope such that the y-axis is perpendicular to the slope & x-axis is positive down the slope.

The only force acting on the system is: $$(m+ M)g\sin\theta = (m + M)a$$. Therefore $$a = g\sin\theta$$.

On the ground, the forces are:

$$-f_k = (M +m)a'$$. Now, normal contact force by the ground $$N = (M + m)g$$. Therefore, $$f_k = \mu_k (m + M)g$$ . Hence, $$a' = -\mu_k g$$.

Now, length of the slope is $\dfrac{h}{\sin\theta}.$ So, the final velocity at the end of the slope is $$v = \sqrt{\left(2g\sin\theta \cdot \dfrac{h}{\sin\theta}\right)} \implies v = \sqrt{2gh}$$. This velocity has direction along the slope and that means the sense of the vector is making angle $\theta$ with the horizontal. Therefore, the horizontal velocity on the ground is $v\cos\theta$(I'm assuming the system's motion from the "centre-of-mass" perspective). Now, while reaching the ground, you know the initial velocity $v$ and also the acceleration $a'$ & final velocity which is zero. Use the equation $${u_{final}}^2 = {u_{initial}}^2 + 2a.s \implies s = \dfrac{{u_{final}}^2 - {u_{initial}}^2}{2a}$$ to find the distance $s$ transversed as you know all the variables(actually the acceleration here is constant assuming $\mu_k$ is same everywhere). I'll leave it for you.

• Sorry, I was busy answering another question. Have you checked the answer from your book? Is it correct? We are restricted not to answer completely the homework quo. But, I assure you, my approach right & even you can use work-energy theorem to reverify your work. If confused, ask anytime for help:)
– user36790
Commented Apr 4, 2015 at 4:31