A child going down a slide with friction

In summary, the slide takes 3.18 seconds to go from the top to bottom of a 10 meter long slope with an inclination of 20 degrees. The child must apply a force of 101.36 Newtons to have the speed of 1.85 kilometers per hour on top of the hill.
  • #1
mmoadi
157
0

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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  • #2
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
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
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
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
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?
 
  • #7
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

  • Forces working on a sled going up incline.bmp
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  • #8
Looks good! I actually got 37.9, but I use g = 9.81 which may well account for it.
 
  • #9
Me too, 37.9 N, but I made a typo:biggrin:
Thank you very much! Have a nice day!:smile:
 

What is friction and how does it affect a child going down a slide?

Friction is the force that resists motion between two surfaces in contact. In the context of a child going down a slide, friction between the child's body and the surface of the slide will slow down their descent and control their speed.

How does the angle of the slide affect the amount of friction experienced by a child?

The steeper the angle of the slide, the greater the friction between the child and the slide. This is because the steeper angle creates a larger normal force, which increases the friction force.

What factors can influence the amount of friction experienced by a child going down a slide?

The amount of friction experienced by a child going down a slide can be influenced by factors such as the material of the slide, the smoothness of the surface, the weight of the child, and the clothing the child is wearing.

Is it possible for a child to go down a slide without any friction?

No, it is not possible for a child to go down a slide without any friction. Even on a very smooth surface, there will always be some amount of friction between the child and the slide due to the microscopic roughness of both surfaces.

How does friction affect the safety of a child going down a slide?

Friction plays an important role in the safety of a child going down a slide. It helps to control their speed and prevent them from sliding too quickly, which could result in injury. However, too much friction can also cause discomfort or burns on the child's skin, so it is important to ensure that the slide is well-maintained and not too rough.

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