Dynamic problem

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



A student is skateboarding down a ramp that is 5.9 m long and inclined at 12° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.3 m/s. Neglect friction and find the speed at the bottom of the ramp.


Homework Equations


This is my steps:
As I dont know the weight of the student, I dont know the force. So I cant use F-ma

Vo=2.3
Vox= 2.3cos78°= 0.48
Voy= 2.3sin78°=2.25
t=0.32


ΔY (found by SOPHATOA) =Voy.t+1/2 a t^2
1.23= 2.25t+1/2(9.8) t^2

ΔX=Vox.t+1/2 a t^2
5.8=0.48(0.32)+1/2.a.(0.32)
ax=110.3

Vx=Vox+at
Vx=0.48+110.3(0.32)
Vx=35.8

Vy=Voy+at
Vy=2.25+9.8(0.32)
Vy=5.386

therefore the resultant of Vx and Vy is 36.2

The Attempt at a Solution



my answer is wrong

I really need some help :D
 

Answers and Replies

  • #2
662
307
Mg is counter-acted by normal force...
For easier methods:
You could just use conservation of energy.
Or you could rotate the coordinate axes.(x-axis along the ramp, y perpendicular)
 

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