Distance Block Slides Up Ramp

  • Thread starter getty102
  • Start date
  • #1
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Homework Statement


A block of mass 2.96 kg is sliding up a ramp with an initial speed of 12.8 m/s. The ramp is inclined from the horizontal at an angle of 36.4 degrees and the coefficient of kinetic friction is [itex]\mu[/itex]=045. What is the displacement of the block along the ramp from the initial time until it stops?


Homework Equations



W=ΔE
W=∫F*ds
ΔE=(m/2)(vf)2-(m/2)(v0)2 where (vf)=0
[itex]\Phi[/itex]=36.4
m=2.96kg
v0=12.8 m/s
[itex]\mu[/itex]= 0.45

The Attempt at a Solution



[itex]\mu[/itex]*m*g*cos[itex]\Phi[/itex]x=-(m/2)(v0)2
solve for x
x=-(m/2)(v0)2/[itex]\mu[/itex]*m*g*cos[itex]\Phi[/itex]
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi getty102! :smile:

correct method, but you forgot to include the gravitational PE (or gravitational work done) :wink:
 
  • #3
38
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Cool, thank you. The potential energy from gravity is dependent on the height. Which in this problem is kind of what I'm looking for because if I can find the height I would know the total displacement of the block up the ramp.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
xsinΦ ? :wink:
 

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