# Determine the speed of the block at the bottom of the inclin

## Homework Statement

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θwith the horizontal. The magnitude of the block's acceleration is gsin(θ) .Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline.
Express your answer in terms of g and the variables m, l, and θ.

UG=−mgsinθ(x−l)

## The Attempt at a Solution

i know that the correct answer does not depend on "m" so i assumed the equation would be v=l/gsinθ. thats inccorrect and i tried reversing it too and still wrong. i know that they won't be multiplied so i'm unsure the relation

gneill
Mentor
What was the "expression you derived in the previous part"? Something to do with velocity, acceleration, and distance perhaps?

What was the "expression you derived in the previous part"? Something to do with velocity, acceleration, and distance perhaps?
this was it : UG=−mgsinθ(x−l)

gneill
Mentor
Okay, well that appears to be an expression for gravitational potential energy. What are x and l in that equation?

You still need to get to velocity somehow. Have you learned about conservation of energy and how kinetic energy and potential energy are related?

Okay, well that appears to be an expression for gravitational potential energy. What are x and l in that equation?

You still need to get to velocity somehow. Have you learned about conservation of energy and how kinetic energy and potential energy are related?
l= inclined plane lenght
x=UG(x=l)=0.
k=-U

gneill
Mentor
l= inclined plane lenght
x=UG(x=l)=0.
k=-U
I have to ask, is UG something different from U, and is k meant to represent kinetic energy?

Do you know how to find the velocity if you have the kinetic energy?

On a related subject, the problem statement tells you that the magnitude of the acceleration is given by gsin(θ). Do you know a standard kinematic equation that relates acceleration, distance, and velocity?

I have to ask, is UG something different from U, and is k meant to represent kinetic energy?

Do you know how to find the velocity if you have the kinetic energy?

On a related subject, the problem statement tells you that the magnitude of the acceleration is given by gsin(θ). Do you know a standard kinematic equation that relates acceleration, distance, and velocity?
Ug is potential energy, k is kinetic energy

Ug is potential energy, k is kinetic energy
yes is i have kinetic energy, velcoity is the square root of k/m

but i don't have kinetic energy, i have potential

would speed be the √mgsinθ(x−l)/0.5m

gneill
Mentor
yes is i have kinetic energy, velcoity is the square root of k/m
Not quite. Since ##k = \frac{1}{2} m v^2##, then solving for v gives ##v = \sqrt{\frac{2 k}{m}}##. Don't forget that "2".

but i don't have kinetic energy, i have potential
But you know how they are related. Once the block moves from the top of the ramp to the bottom, what happens to the potential energy? Where does the energy go?

would speed be the √mgsinθ(x−l)/0.5m
It's hard to tell because I can't be sure what the square root sign is meant to cover. And I'm still not sure what x is meant to be. I understand that l is the inclined plane length from post #5:
l= inclined plane lenght
x=UG(x=l)=0.
k=-U
But I'm not sure how to read your second equation. Surely you're not equating a length to a potential energy?

The square root sign is meant to cover everything

gneill
Mentor
The square root sign is meant to cover everything
In that case I believe that it will yield the velocity. Note that the mass m cancels in the numerator and denominator. It would probably be more clear if you were to replace (x - l) with a single variable like "d" for the distance traveled along the slope.

Note that there's also a kinematic equation that you should be familiar with that would get you to the result quickly if you know the acceleration and distance (which you do in this case).