A problem on two dimensional kinematics

[Winsy]

Homework Statement

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?

The answer is 3.73m.
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?

 

[Winsy]

This is my attempts.

 

[Winsy]

I can't read your writing but a MUCH easier way is to use conservation of energy. Think: at the top, the block is not moving, so all its K.E is converted to gravitational P.E

[Moderator: complete solution provided by DeldotB has been removed]

But why we can use the velocity directly without resolving it into x and y directions? And since my classes haven't covered the conservation of energy, I would like to see what's going wrong with my answers.

 

[Winsy]

But why we can use the velocity directly without resolving it into x and y directions? And since my classes haven't covered the conservation of energy, I would like to see what's going wrong with my answers.

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.

 

[Winsy]

I can't read your writing but a MUCH easier way is to use conservation of energy. Think: at the top, the block is not moving, so all its K.E is converted to gravitational P.E

$1/2mv^2=mgh$ the m's cancell, so you have $v^2/2=gh$ which implies $h=v^2/(2g)$ Sine = opposite/hypotenuse, so $sin(20)=h/d$ so $h=sin(20)d$ substituting that into the above eqn, $v^2/(2g)=sin(20)d$ so $d=v^2/(2gsin(20))=3.729 m$

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.

 

[Nathanael]

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

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.