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Potential energy -- child sleds down a slope and comes to a stop...

  • Thread starter eefje
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  • #1
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Homework Statement


A child starts from rest and slides on a sled first slope
down over a distance of 25 m . Then, the child slides further over a
horizontal plane. The angle of inclination is 35 ° and the kinetic
coefficient of friction between the sled and the snow is 0.100. The mass
of the child and the slide together 23 kg . What distance puts the child at the
horizontal section off before coming to a halt ?(122,93) What do I wrong? thank you

Homework Equations


dE=-Fw.d Ek=(1/2m).v^2 Epot= m.g.h


The Attempt at a Solution


first I go from the begining to the ground then second I start from the ground to when it stops
y=-cos35.G+N=0
N=23*9,81N*cos(35°)
= 184,8
Fw= Fn*0,1= 18,48
x=-cos55*23*9,81+18,48=m*a
-110=m*a

dE=-Fw*d
m*g*hB+0-(1/2m*v^2A)-0=-Fw*d

(hB= cos55°*25=14,34)

23*9,81*14,34-1/2*23*v^2A=-18,48*d
v^2A=321


What distance puts the child at the
horizontal section off before coming to a halt
-1/2*m*v^2A=-Fw*d
-1/2*23*321+18,48=d
 
Last edited:

Answers and Replies

  • #2
963
213

Homework Statement


A child starts from rest and slides on a sled first slope
down over a distance of 25 m . Then, the child slides further over a
horizontal plane. The angle of inclination is 35 ° and the kinetic
coefficient of friction between the sled and the snow is 0.100. The mass
of the child and the slide together 23 kg . What distance puts the child at the
horizontal section off before coming to a halt ?(122,93) What do I wrong? thank you

Homework Equations


dE=-Fw.d Ek=(1/2m).v^2 Epot= m.g.h


The Attempt at a Solution


first I go from the begining to the ground then second I start from the ground to when it stops
y=-cos25.G+N=0
N=23*9,81N*cos(25°)
= 204,5N
Fw= Fn*0,1= 20,45
x=-cos65*23*9,81+20,45=m*a
-74,9=m*a

dE=-Fw*d
m*g*hB+0-(1/2m*v^2A)-0=-Fw*d (hB= cos65°*25=10,57)
23*9,81*10,57-1/2*23*v^2A=-20,45*d
v^2A=252


What distance puts the child at the
horizontal section off before coming to a halt
-1/2*m*v^2A=-Fw*d
-1/2*23*252+20,45=d
kindly recheck your calculation to see that it is correct- angle of inclination is 35 degree- but you are using 25 degrees.

you could have used Potential energy (less workdone against friction) to calculate velocity square (final) and the calculared d straight away as K.E. lost against work done against friction.
 
  • #3
15
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Thank you very much !! I have changed the numbers now of the degrees but I don't understand what I have to do now
 
  • #4
berkeman
Mentor
56,806
6,773
A child starts from rest and slides on a sled first slope
down over a distance of 25 m . Then, the child slides further over a
horizontal plane. The angle of inclination is 35 ° and the kinetic
coefficient of friction between the sled and the snow is 0.100. The mass
of the child and the slide together 23 kg . What distance puts the child at the
horizontal section off before coming to a halt ?
I could be wrong, but I don't think the velocity comes into this problem. You have the initial PE of the child and sled, and that equals the work done by friction against the sled's motion along the path down the slope and on the flat. The normal force is different on the slope versus on the flat, but that is the only thing you should have to change for the two parts of the path, IMO.

Can you try to do the problem just based on the initial PE and the work done by friction along the path?
 
  • #5
963
213
well,

the loss in P.E. must have geenerated the kinetic energy but the friction on the incline must have dissipated part of the P.E. so till the cart comes to the horizontal end
write the energy equation-
P.E. - Work done against Friction= K.E. of the cart with boy
now equate the K.E. found above with the Energy loss in going against Friction during Translation distance d on horizontal till it stops
the mass of boy and cart may not be needed.
 
  • #6
15
0
I could be wrong, but I don't think the velocity comes into this problem. You have the initial PE of the child and sled, and that equals the work done by friction against the sled's motion along the path down the slope and on the flat. The normal force is different on the slope versus on the flat, but that is the only thing you should have to change for the two parts of the path, IMO.

Can you try to do the problem just based on the initial PE and the work done by friction along the path?
Thank you! you mean just:
m*g*h=-Fw*d
23*9,81*14,3=18,48
d=175,08
it isn't right
is mine hB right?
 
  • #7
berkeman
Mentor
56,806
6,773
Thank you! you mean just:
m*g*h=-Fw*d
23*9,81*14,3=18,48
d=175,08
it isn't right
is mine hB right?
There are different normal forces on the slope versus the flat, so the friction force is different on the slope versus the flat. It doesn't look like you've taken that into account yet... :smile:
 
  • #8
15
0
There are different normal forces on the slope versus the flat, so the friction force is different on the slope versus the flat. It doesn't look like you've taken that into account yet... :smile:
Is there only friction on the flat?. I have put the friction of the slope =0 and changed everything in the thing
I have post and I come still on 174,86
 
  • #9
berkeman
Mentor
56,806
6,773
Is there only friction on the flat?.
No, there is snow everywhere, so there is friction everywhere. The friction force is a bit lower on the slope, though, since the normal force is smaller. How much smaller?
 
  • #10
15
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No, there is snow everywhere, so there is friction everywhere. The friction force is a bit lower on the slope, though, since the normal force is smaller. How much smaller?
Thank you :) but where is the friction 0,1 then? I will try to solve it an other time because I have tried it and i can't solve it maybe another time with the help of the things you have said
 
  • #11
berkeman
Mentor
56,806
6,773
but where is the friction 0,1 then?
That is the coefficient of friction between the sled and the snow. μ = 0.1 everywhere between the sled and the snow.

When you listed the Relevant Equations, you did not list the equation that relates the friction force F to the normal force N and the coefficient of friction μ. Can you list that equation? What is N on the slope and on the flat? N is different on the slope versus the flat...
 
  • #12
15
0
well,

the loss in P.E. must have geenerated the kinetic energy but the friction on the incline must have dissipated part of the P.E. so till the cart comes to the horizontal end
write the energy equation-
P.E. - Work done against Friction= K.E. of the cart with boy
now equate the K.E. found above with the Energy loss in going against Friction during Translation distance d on horizontal till it stops
the mass of boy and cart may not be needed.
Thank you :smile:
 
  • #13
15
0
well,

the loss in P.E. must have geenerated the kinetic energy but the friction on the incline must have dissipated part of the P.E. so till the cart comes to the horizontal end
write the energy equation-
P.E. - Work done against Friction= K.E. of the cart with boy
now equate the K.E. found above with the Energy loss in going against Friction during Translation distance d on horizontal till it stops
the mass of boy and cart may not be needed.
Thank you I understand it now
 
  • #14
15
0
That is the coefficient of friction between the sled and the snow. μ = 0.1 everywhere between the sled and the snow.

When you listed the Relevant Equations, you did not list the equation that relates the friction force F to the normal force N and the coefficient of friction μ. Can you list that equation? What is N on the slope and on the flat? N is different on the slope versus the flat...
Thank you very much I understand it now :smile:
 
  • #15
15
0
I have tried it with
Epot=-Fw.d with TWo different FN but the solution is wrong
 
  • #16
berkeman
Mentor
56,806
6,773
I have tried it with
Epot=-Fw.d with TWo different FN but the solution is wrong
Well let me try to read your mind to see where the problem is.... :smile:

But seriously, please post your work in detail so we can try to find any errors...
 
  • #17
15
0
N1=23*9,81N*cos(35°)
= 184,8
Fw1= Fn*0,1= 18,48
N2=23*9,81=225,63
Fw2=Fn*0,1=22,563
(hB= cos55°*25=14,34)
Epot=-Fw.d
23*9,81*14,34=-18,48*25-22,563*d
3235,53=-462-22,563*d
3235,53+462=-22,563*d
3697,53/-22,563=d

163,87=d
 
  • #18
haruspex
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Epot=-Fw.d
Be careful with signs. With that formulation, the Epot should be the change in potential energy, yes? Is that positive or negative?

I strongly encourage you to get into the habit of working purely symbolically as far as possible. Pretend no numeric values are given, so make up variables to represent them. It has many benefits. In the present case, all the mass terms and g terms would have cancelled, simplifying the arithmetic at the end. You would also get more precise answers numerically since you would avoid accumulation of rounding errors. See https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/ for more on the subject.
 
  • #19
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But how do i come to the Right solution ? What did i do Wrong in the latest answer i gave, and not only the positive of negative sign
 
  • #20
haruspex
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But how do i come to the Right solution ? What did i do Wrong in the latest answer i gave, and not only the positive of negative sign
As far as I can see, the signs are your only errors.
 
  • #21
Total PE gets dissipated by total frictional force (along incline + along horizontal path) and KE becomes zero at rest.

So, PE = Wfr1 + Wfr2, solve for Dhorizontal.
 
  • #22
haruspex
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Total PE gets dissipated by total frictional force (along incline + along horizontal path) and KE becomes zero at rest.

So, PE = Wfr1 + Wfr2, solve for Dhorizontal.
That's what @eefje did in post #17, but had a sign error or two.
 

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