# A child going down a slide with friction

1. Oct 28, 2009

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

A child slides with a slide from a summit of a 10m long slope and with an inclination of 20º. Coefficient of kinetic friction between the slide and the slope is 0.15.
- How much time does it take the slide to go from the top to the bottom?
- With how much permanent force upwards must the child push the slide up the same slope for the slide to have the speed of 1.85 km/h on top of the hill?
The mass of the slide is 8kg. The child is pushing the slide with a force that is parallel to the slope.

2. Relevant equations

ΣF(x)= mg*sinθ – mg*cosθ* µ(k)= ma(x) → a(x)= g(sinθ – cosθ*µ(k))
ΣF(y)= N- mg*cosθ= 0
x(final)= x(initial) + v(x-initial) + ½ a(x)*t²

3. The attempt at a solution

- How much time does it take the slide to go from the top to the bottom?
d= x(final) = 10 m
θ= 20º
µ(k)= 0.15
x(initial)= 0
v(x-initial)= 0

x(final)= x(initial) + v(x-initial) + ½ a(x)*t² → d= ½ a(x)*t²= ½g(sinθ – cosθ*µ(k))*t²
d= ½g(sinθ – cosθ*µ(k))*t² → t= sqrt(2d / g(sinθ – cosθ*µ(k))
t= 3.18 s

Is this part correct?

What about the second part? How do I approach it? Any hints? Thanks!

2. Oct 28, 2009

### Delphi51

Yes, I get the 3.18 too.
The second part is kind of the reverse - you begin with calculating the acceleration from the known Vf and work your way back to the forces.

3. Oct 28, 2009

Second part:
- With how much permanent force upwards must the child push the slide up the same slope for the slide to have the speed of 1.85 km/h on top of the hill?
The mass of the slide is 8kg. The child is pushing the slide with a force that is parallel to the slope.

Equations:

v(final) ²= v(initial) ² + 2ax
N= m(slide)g*cosθ + m(child)g*sinθ
F(net)= ma
F(k)= µ(k)*N
F(child)= m(child)g*cosθ
F(slide)= m(slide)g*sinθ

First the acceleration:
1.8 km/h= 0.5 m/s
v(final)²= v(initial) ² + 2ax
a= 0.0125 m/s²

I named [m(child)g]= F and obtained:
F(net)= F*cosθ – m(slide)g*sinθ - µ(k)*m(slide)g*cosθ + µ(k)*F*sinθ

F= m(slide)*a + m(slide)g*sinθ + µ(k)*m(slide)g*cosθ / cosθ + µ(k)*sinθ
F= 101.36 N

Are my calculations correct?
I hope I did it right!

4. Oct 28, 2009

### Delphi51

I get a slightly larger number for the acceleration because I used 1.85 km/h instead of 1.8 km/h. But quite a difference in the force calc.
I don't understand the "/ cosθ + µ(k)*sinθ" at the end of it. What is that for? The last term doesn't even have the units of force.

5. Oct 28, 2009

This is what I did:

F(net)= F*cosθ – m(slide)g*sinθ - µ(k)*m(slide)g*cosθ + µ(k)*F*sinθ
m(slide)*a= F*cosθ – m(slide)g*sinθ - µ(k)*m(slide)g*cosθ + µ(k)*F*sinθ
m(slide)*a + m(slide)g*sinθ + µ(k)*m(slide)g*cosθ = F*cosθ + F*µ(k)*sinθ
m(slide)*a + m(slide)g*sinθ + µ(k)*m(slide)g*cosθ = F(cosθ + µ(k)*sinθ)
F= m(slide)*a + m(slide)g*sinθ + µ(k)*m(slide)g*cosθ / (cosθ + µ(k)*sinθ)

And after putting in all the information, I got the result F= 101.36 N.

Where did I went wrong?

6. Oct 28, 2009

### Delphi51

Why is F multiplied by cosθ? It says the child pushes parallel to the slope, so no cos is called for. The last term doesn't make sense to me at all. What opposing force could there be besides gravity and friction?

7. Oct 28, 2009

I corrected the acceleration calculation:
1.85 km/h= 0.531 m/s and a= 0.0132 m/s²

I think I complicated a simple problem, so I started all over.
First I drew a diagram of forces (see picture).

Horizontal component:
F(x)= F(g)= mg sin θ

Vertical component:
F(y)= N= mg cos θ

Frictional force:
F(f)= µ(k) N

Sum of all forces working on the sled:
F(net)= ma= F(applied) - F(f) - F(g)

F(applied)= ma + F(f) + F(g)
F(applied)= 37 N

Did I solve it correctly this time?
Thanks for helping!

#### Attached Files:

• ###### Forces working on a sled going up incline.bmp
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8. Oct 28, 2009

### Delphi51

Looks good! I actually got 37.9, but I use g = 9.81 which may well account for it.

9. Oct 29, 2009