Force of Friction on an Inclined Plane

  • Thread starter mmalone11
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  • #1
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Homework Statement


The diagram shows a 5kg block of lead released from rest at the top of an incline. The block has a speed of 6 m/s when it reaches the bottom. The angle between the slope and the ground is 40° and the slope is 10 m long.
a) What is its PE at the top?
b) What is its KE at the bottom?
c) What is the work done by friction?
d) What must be the coefficient of friction?


Homework Equations


I am having trouble finding the friction. Once i find the friction, i know how to find the coefficient force of friciton using Ffr=mu*Fn.

The Attempt at a Solution


For part a first I found what the height was by doing 10(sin(40))=Height and got 6.43 m. Than i plugged that into PE=mgh... 5(9.8)(6.43) ... getting 315.07J ...
For part b I used KE=1/2(m)(v2) ... 1/2(5)(62) ... getting 90J ...
 

Answers and Replies

  • #2
kuruman
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So you start with with 315 J of potential energy and zero kinetic energy. How much total energy would you have at the bottom of the incline if there were no friction?
 
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  • #3
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Would that still be 315 J?
 
  • #4
kuruman
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Correct. With friction you only have 90 J. Where did the rest of the Joules go?
 
  • #5
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Heat is the only thing i could think of? But how do you know 90 J went to friction
 
  • #6
kuruman
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I never said 90 J went to friction. At the bottom of the incline, you have 90 J. This means that 315-90 = 225 J went somewhere. As you say it is heat and this heat is generated by friction. So the block traveled 10 m and lost 225 J to friction. How can you use this information to find the work done by friction?
 
  • #7
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Well i know Work=Fd but im not sure how to incorporate the joules..
 
  • #8
kuruman
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The units of both work and energy are Joules.
 
  • #9
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so just divide by 10?
 
  • #10
kuruman
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Divide by 10 to get the answer to what part?
 

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