Calculating Velocity of a Sliding Box on an Inclined Ramp

[Cglez1280]

Homework Statement

A box of mass 6.2 kg slides 4.8 m down a ramp that is inclined 38 degrees to the horizontal. What is the velocity of the box right before it hits the ground?
Mass=6.2 Height=dsintheta? V=is what needs to be found theta=38

Homework Equations

GPE=mgh=mgdsin(theta)
KE=1/2mv^2
New equation:
GPE⋅+KE⋅=GPE
mgh⋅+1/2mv^2=mgh
1/2mv^2=mgh-mgh⋅
mv^2=2(mgh-mgh⋅)
V^2=2(mgh-mgh⋅)/m
v=√2(mgh-mgh⋅)/m

The Attempt at a Solution

i just want to see if my equations are correct. if it could be factored even more could someone tell me. like the two mgh's i feel like could be changed but I am not too sure. I know h is also h=dsin(θ) but would it be the same for the two since it will be different?

 

[PeroK]

I'm not sure I understand what you are trying to do. I think you're trying to say something like:

The box falls a vertical height $h = d\sin(\theta)$ where $d$ is the length of the slope.

The loss of GPE is, therefore, ...

 

[Cglez1280]

I'm not sure I understand what you are trying to do. I think you're trying to say something like:

The box falls a vertical height $h = d\sin(\theta)$ where $d$ is the length of the slope.

The loss of GPE is, therefore, ...

I'm sorry that's what I'm trying to say, I just need a little bit of help on this problem

 

[PeroK]

I'm sorry that's what I'm trying to say, I just need a little bit of help on this problem

Can you do the next step? What is the loss in GPE?