Children's Slide, given slide length and angle, solve speed at bottom

1. Sep 7, 2014

speedtriple

1. The problem statement, all variables and given/known data

Hi guys! Here is my problem:

A Children's slide is 20 ft long and makes an angle 30° with the horizontal. If the coefficient of sliding friction is 0.05 and the child starts from rest at the top, with what speed does he reach the bottom?

I want to say it is about potential energy? or is it about the work of nonconservative forces? I appreciate any input! thank you!

2. Relevant equations

noncoservative Work = Change in K + Change in U

3. The attempt at a solution

2. Sep 7, 2014

voko

This can be solved in more than one way. But using the work-energy principles is probably easiest.

3. Sep 7, 2014

speedtriple

As in the Net work is equal to change in kinetic Energy? I understand using that but what throws me off is that there is no mass to use it with the equation. I'm thinking it is just kept as the variable?

4. Sep 7, 2014

voko

Keep the mass as a symbolic variable m. If the problem is posed correctly, it will disappear from equations at some point.

5. Sep 7, 2014

speedtriple

SO how I see it, is It is a 30-60-90 triangle with gravity acting on the Child on top of the slide. Since it asks what is at the bottom, the height is 0ft or the velocity initially is 0 which makes the equations I use become zero. I'm sorry if this is really wrong but I'm getting really confused here..

6. Sep 7, 2014

voko

Potential energy at the top? At the bottom?
Kinetic energy at the top? At the bottom?
Are the any non-conservative forces involved? What is their work?

7. Sep 7, 2014

speedtriple

Well when I get the potential energy at the top, I use Ug=mgy. So Ug= (m)(9.8)(10). I got 10 since the ladder is 20 ft long and the incline is 30 degree which makes the initial heigh 10. The kinetic energy at the top would be 0 since he's not moving? So how do I get the change in kinetic energy if I am subtracting K1 from k2? The non conservative forces friction which relates to the work being done on the object but Im not sure how I'm going to solve for work without mass to get the force with the displacement. Is this problem really solvable or maybe it needs as mass to be solved?

8. Sep 7, 2014

voko

The general approach is correct in the above. However, can you really mix 9.8, which is meters per seconds squared, with 10, which is feet?

Just like with the potential energy above, use 'm' for mass.

Note you have only answered two of my questions, and even those not entirely.

9. Sep 7, 2014

speedtriple

AHA! sneaky sneaky tips..

So I found the change in gravitational potential energy = -(m)29.8704
and the change in Kinetic Energy = .5(m)Vfinal^2

I'm looking at my notes and it seems there is this equation for the conservation of Mechanical Energy

change in Kinetic + Change in Potential = 0

I plugged in the change in Ug and Change in K and set them equal to 0

You're Right! The m cancelled out and then I solved for Vfinal via algebra! Equals 7.7 m/s???

oh lordy I hope this is right!

Last edited: Sep 7, 2014
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