Horizontal Plane With Friction And Incline Plane With no Friction

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  • #1
amgoda
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1. A body with mass m = 5 kg and initial velocity = 5 m/s moves along a horizontal plane with length 10 m, where the friction coefficient is μ=0.1, After passing the part with friction, the body moves up along a tilted plane which has no friction. What is the maximum height of body elevation along the tilted plane?

http://imageshack.us/photo/my-images/15/physicso.png/

i tired to get the final velocity before it goes up the inclined plane and got (2.32) but i think i got it wrong
 

Answers and Replies

  • #2
Dick
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You should really show how you got 2.32m/s. And no, you didn't get it wrong. But you never really needed to find the velocity. What you needed to find is the kinetic energy before it goes up the ramp. How is that related to the height it climbs?
 
  • #3
amgoda
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alright .. here is how i thought about it ..
N=mg
Fκ=μk.N
a=(μk.mg)/-m=-.98m/s^s
vf^2=V0+2a(x-x0)
vf^2=5^2+(2*-.98*10)= 2.32
and i thought that the final velocity on the horizontal plane is the initial velocity for the inclined plane ..
 
  • #4
Dick
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alright .. here is how i thought about it ..
N=mg
Fκ=μk.N
a=(μk.mg)/-m=-.98m/s^s
vf^2=V0+2a(x-x0)
vf^2=5^2+(2*-.98*10)= 2.32
and i thought that the final velocity on the horizontal plane is the initial velocity for the inclined plane ..

That is perfectly fine. I don't think it's the most economical route to the answer but you are doing very well. But now how do you use the initial velocity up the ramp to find the height you climb?
 
  • #5
amgoda
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that's the problem .. i just got to this point and i don't know what to do afterwards ...
btw thanks for the fast replies
 
  • #6
Dick
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that's the problem .. i just got to this point and i don't know what to do afterwards ...
btw thanks for the fast replies

Do you know how to use energy conservation to do it? Gravitational potential energy at a height h up the ramp is mgh. That must be equal to the kinetic energy you started with. Yes?
 
  • #7
amgoda
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ahaa .. so it should equal to the kinetic energy at the end of the horizontal plane .. so will it be .5*m*v^2=mgh .5*5*2.32^2=5*9.8*h
h=0.63 meters ?
 
  • #8
Dick
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ahaa .. so it should equal to the kinetic energy at the end of the horizontal plane .. so will it be .5*m*v^2=mgh .5*5*2.32^2=5*9.8*h
h=0.63 meters ?

Exactly, except I don't get h=0.63m. Can you try that once more? And think about whether you really needed to find v at all. You just need the KE at the start of the ramp. Can you think of a simpler way to find that?
 
Last edited:
  • #9
amgoda
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ahaaa .. here we go .. i don't know why i haven't thought of this
Wnet=μk.mg.Δx = change in KE
then KE at the bottom = 13.5 =mgh
h= .027
and i recalculated H in the First method i used and i got 0.27 also ..
thanks a lot u drew my attention to the easier way .. thanks
 
  • #10
Dick
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ahaaa .. here we go .. i don't know why i haven't thought of this
Wnet=μk.mg.Δx = change in KE
then KE at the bottom = 13.5 =mgh
h= .027
and i recalculated H in the First method i used and i got 0.27 also ..
thanks a lot u drew my attention to the easier way .. thanks

Very welcome. Energy conservation is often the easy way to go. And both are h=0.27, right?
 
  • #11
amgoda
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yup.. i just messed up in the first calculation
 

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