Homework Help: A problem on two dimensional kinematics

1. Oct 15, 2015

Winsy

1. The problem statement, all variables and given/known data

A block is given initial velocity of 5m/s up a frictionless 20 degrees incline. How far up the incline does the block slide before coming to rest?

I set the coordinates to be perpendicular to the plane instead of parallel to the incline. I calculated the vertical and horizontal displacement and then use Pythagorean Theory to evaluate how far the car goes. But I did not get the expected answer, which is 3.73. Can anybody tell me what's wrong with my solusions?

2. Oct 15, 2015

Winsy

This is my attempts.

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3. Oct 15, 2015

Winsy

Last edited by a moderator: Oct 16, 2015
4. Oct 15, 2015

Winsy

v(y0)=v(0)sin20°=5m/s×sin20°=1.71m/s
since v(t)^2-v(y0)^2=2as,v(t)=0,a=-9.8m/s^2
h=v(y0)^2/2g=0.149
since a=(v(t)-v(0))/t ,t=v(t)-v(y0)/g=0.174s

v(x0)=v(0)×cos20°=4.7m/s
x=(v(xo)+v(t))×t/2=0.409m
s=√[x^2+h^2]=0.435m

That's what I've tried.

5. Oct 16, 2015

Winsy

v(y0)=v(0)sin20°=5m/s×sin20°=1.71m/s
since v(t)^2-v(y0)^2=2as,v(t)=0,a=-9.8m/s^2
h=v(y0)^2/2g=0.149
since a=(v(t)-v(0))/t ,t=v(t)-v(y0)/g=0.174s
v(x0)=v(0)×cos20°=4.7m/s
x=(v(xo)+v(t))×t/2=0.409m
s=√[x^2+h^2]=0.435m
That's what I've tried.

6. Oct 16, 2015

Nathanael

a is not -9.8 m/s^2 in this situation, because the normal force also has a vertical component.

I would suggest resolving gravity into it's components along the incline and perpendicular to the incline. Then work with the entire magnitude of velocity.