A child going down a slide with friction

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Homework Help Overview

The discussion revolves around a physics problem involving a child sliding down a slope with friction. The problem includes calculating the time taken to slide down a 10m slope inclined at 20º and determining the force required for the child to push the slide back up to achieve a specific speed at the top.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of time taken to slide down and the necessary force to push the slide back up. There are attempts to derive equations of motion and forces acting on the slide, with some participants questioning the validity of their calculations and the assumptions made.

Discussion Status

Some participants agree on the time calculation, while others explore different approaches to the force calculation. There is an ongoing examination of the equations used and the components of forces involved, with no explicit consensus reached on the final calculations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information available for discussion. There are also questions regarding the setup of forces and the interpretation of the problem, indicating potential ambiguities in the problem statement.

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Homework Statement



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.

Homework 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

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!
 
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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.
 
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!
 
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.
 
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?
 
F(net)= F*cosθ – m(slide)g*sinθ - µ(k)*m(slide)g*cosθ + µ(k)*F*sinθ
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?
 
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!
 

Attachments

Looks good! I actually got 37.9, but I use g = 9.81 which may well account for it.
 
Me too, 37.9 N, but I made a typo:biggrin:
Thank you very much! Have a nice day!:smile:
 

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