# Spring and box on incline

[SOLVED] spring and box on incline

## Homework Statement

A 50 kg mass rests atop a 20m high incline. The spring constant of a 10m long spring below is 1000N/m. The 50 kg crate is sent down the incline. Assume all surfaces in this problem are frictionless.
What is the velocity of the crate traveling along the horizontal surface?

Fspring=kx
F=ma

## The Attempt at a Solution

I am so confused by this problem. I tried to find the length of the incline, but couldn't figure out how. The only thing I could think to do was to find the potential energy of the crate, which is 9810 J. I know that only force gravity is acting of the crate, and that Fgy=FN, but that doesn't seem to help me at all.
Please help me if you can because I have a physics final exam tomorrow. Thank you so much.

## The Attempt at a Solution

#### Attachments

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Use energy...

Doc Al
Mentor
Assume all surfaces in this problem are frictionless.
Since there's no friction, what's conserved?

I don't want you to give me the answer, but is there anyway you could give me a little more information? I know that energy should be conserved, but the only formulas I know for the conservation of energy involve velocity, which I can't figure out a way to find.

Doc Al
Mentor
There are three types of energy involved here: kinetic energy, gravitational potential energy, and spring potential energy. Only kinetic energy involves velocity.

Is this right? :

1/2mv^2 + mgy = 1/2mv^2 + mgy
0 + (50kg)(9.81m/s^2)(20m) = 1/2(50kg)v^2 + 0
19.81m/s = v

So then the spring would be compressed .9905m because:
f = ma
f = 50kg(19.81m/s)
f = 990.5N

fp = kx
990.5N = 1000N/m * x
.9905m = x

-KAT444

Doc Al
Mentor
Is this right? :

1/2mv^2 + mgy = 1/2mv^2 + mgy
0 + (50kg)(9.81m/s^2)(20m) = 1/2(50kg)v^2 + 0
19.81m/s = v
This is exactly correct. So then the spring would be compressed .9905m because:
f = ma
f = 50kg(19.81m/s)
f = 990.5N

fp = kx
990.5N = 1000N/m * x
.9905m = x
This is not correct.

To find out how the spring compresses, again consider energy. What's the formula for the energy of a compressed spring?

Is this the right way:

elastic PE = 1/2kx^2
energy is conserved so . . .
9810J = 1/2(1000N/m)(x^2)
4.43m = x

Thanks - KAT444

Doc Al
Mentor
Is this the right way:

elastic PE = 1/2kx^2
energy is conserved so . . .
9810J = 1/2(1000N/m)(x^2)
4.43m = x
Perfecto!

I agree with your profile, you're an all around good guy.

Thanks again - KAT444