Solving for the Height of a Slope with Unknown Mass

AI Thread Summary
To find the height of a slope with an unknown mass and no friction, energy conservation principles can be applied, equating kinetic energy at the bottom to potential energy at the top. The equations KE = mv²/2 and PE = mgh indicate that mass cancels out, allowing height to be determined without knowing the mass. The acceleration of the box down the slope can be derived from the gravitational force component acting along the slope. Drawing a free body diagram may also help visualize the forces involved. Understanding these concepts is crucial for solving the problem effectively.
Ivar
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Homework Statement



A box is on top of a slope. The slope has NO FRICTION. V0 = 0 m/s, V = 4m/s.

MASS IS UNKNOWN

What is the height of the slope?

Homework Equations


Equation for energy (i think)

The Attempt at a Solution


I attempted to find the accelleration, but I do not know what the accelleration is for an object running down a slope with no friction.
The only force applying to the box is gravity, I think.

I am new to Physics. I tried solving it, but to no avail.
I know this can be solved without knowing the mass, but I don't know how! Help?
 
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KE at bottom should be equal to PE on the top of the slope
 
Yes. That I know. Bit how does this help me find the height of the Slope?

And what is the accelleration of an object running down a slope with no friction?
 
KE= mv^2 / 2 = mgh =PE .
mass is omitted in the equation , therefore you can find h .
the acceleration = F / m , you should find which component of gravity actually makes the mass running down the slope , then divide m , you got acceleration
 
Hi Ivar. Welcome to Physics Forums.

Jake Jungle has shown one method of getting the answer to your problem using energy conservation. But, it is not the only method, and maybe you are not that comfortable with using energy conservation. If not, have you considered drawing a free body diagram, and showing the forces acting on the box in your free body diagram?

Chet
 

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