# Comparing Acceleration and Time Of Different Slopesm

1. Apr 29, 2016

### dlacombe13

1. The problem statement, all variables and given/known data
There are three slides, slides A, B, and C. There is no friction. The heights are all the same. The masses of each person (initial location at the top of the slide, h) are ranked: mA < mB < mC. Compare each of the following quantities of the sliders using <, >, or =.
a) Initial potential energy
b) Final kinetic energy
c) Final speeds
d) Magnitudes of final acceleration
e) total distance traveled
f) Total time to reach the bottom

2. Relevant equations
mgy = (0.5)mv2
P.E=mgy
K.E=(0.5)mv2

3. The attempt at a solution
I believe so far that:
a) A<B<C
b) A<B<C
c) A=B=C
e) I have to ask if he means displacement along x-axis or literally along the path

So for d and f, I am a little stuck. I think that it might involve the equation ∑F = Δp / Δt. I am pretty sure the only forces acting on the riders are their weight (mg) and the normal force (n). I get the equation:
n-mg = (mV - 0) / t . When I try test values for V (same for all scenarios) and m, I end up with two unknowns, n and t, so I can't seem to reach any conclusions. Any help?

2. Apr 29, 2016

### drvrm

n-mg = (mV - 0) / t . this equation is not correct as the direction of n and mg are not same and not opposed to each other ...one should draw force diagram and see the net force along the incline.

Last edited: Apr 29, 2016
3. Apr 29, 2016

### dlacombe13

Okay that does make sense. I think I'm confused because for some reason my mind wants to tell me that since slope A is more gradual, it will take more time for it to accelerate to the same speed. I can't seem to wrap my mind around it. Can you explain why this is not the case?

4. Apr 29, 2016

### drvrm

initially i missed your diagrams ; i thought the three slides being identical .
so i later edited my response...
regarding time taken by the body to reach the base - its acceleration varies as the curvature of the path changes ..if you draw a free body diagram the acceleration along the path is proportional to sine of theta,where angle is the slope of the path ... so where it is more steep it will be larger sin of angle move from zero to one as angle goes from 0 to 90.
but as the final velocity is same the average acceleration may not be same .. i am just thinking aloud. so the time taken should not be same.; it must depend on path length.

5. Apr 29, 2016

### dlacombe13

Ah, so I was thinking along the right path. I see how the normal force is wrong because, since gravity acts downwards and the angle of the surface (and the rider) would change. So if my logic is correct, slide A would have the greatest FINAL acceleration since the end of the slide has a nearly vertical slope, slide B would have the smallest final acceleration since the end slope is zero, and slide B would be in the middle since it is diagonal, and has a slope in the middle of A and C.
Now for the time, I think I can assume that slides A and C have the same length of path, but flopped over the x-axis. However, I feel like slide C will take less time since it starts out faster. I can't, however, seem to come up with the logic to compare the straight slide with the curved ones...any idea? I feel like it might be less than slide A but more than slide C.

6. Apr 29, 2016

### drvrm

yes , you are thinking along right direction - the weight acts vertically down the normal reaction is along normal to contact surface so it will be normal to the path length;
if the path is making an angle theta with the horizontal then the net force along the path will be mgsin(theta) so acceleration will be F/m = g.sin of theta.
so if the theta is larger-acceleration will be larger.
regarding time taken by the body the c-slide will take the least time as acceleration for the most of its path will be larger.; next comes A slope as far as time taken is concerned and last will be B-slide.
regarding the least time for c ,there are other info is that -as its closer to cycloid curve and that takes least time .

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