Spring and box on incline

  • Thread starter KAT444
  • Start date
  • #1
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[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?

Homework Equations


V_(f^2 )=V_(i^2 )+2aD
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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
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Use energy...
 
  • #3
Doc Al
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Assume all surfaces in this problem are frictionless.
Since there's no friction, what's conserved?
 
  • #4
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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.
 
  • #5
Doc Al
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There are three types of energy involved here: kinetic energy, gravitational potential energy, and spring potential energy. Only kinetic energy involves velocity.
 
  • #6
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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

Thanks for all your help

-KAT444
 
  • #7
Doc Al
Mentor
45,250
1,601
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. :smile:

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?
 
  • #8
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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
 
  • #9
Doc Al
Mentor
45,250
1,601
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!
 
  • #10
6
0
I agree with your profile, you're an all around good guy.

Thanks again - KAT444
 

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