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Velocity for a puck up a ramp

  • Thread starter talaroue
  • Start date
  • #1
303
0

Homework Statement


What minimum speed does a 136.4 g puck need to make it to the top of a 3.31-m-long, 24° frictionless ramp?


Homework Equations


K=mV^2/2 U=mgY



The Attempt at a Solution


PHysics9.jpg
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
the 'y' in your equation is the vertical distance, not horizontal. So the cos24 should be sin24
 
  • #3
wrong its cos.

cos angle = b/c

f * s + potential energy = mv^2 / 2

edit: mistake if you want height then its sin.
 
  • #4
if the equation is mgh = mv^2/2 then you can cancel out the mass.

gh = v^2/2

9.82 * 1.35m = v^2 / 2

26.5 =v^2

v = 5.15m/s

but it think this is totally wrong.
 
  • #5
Cyosis
Homework Helper
1,495
0
You have the angle, hypotenuse and want to know the opposite side. Therefore you need to use the sine.
 
  • #6
303
0
ahhhhh. Thanks guys i see where i went wrong you guys are awesome!!
 

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