How to find Potential Energy without distance

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SUMMARY

The discussion focuses on calculating potential energy and distance traveled by a cube-block system after a collision. The cube, with a mass of 2 kg and speed of 20 m/s, collides with a 5 kg block attached to a spring with a spring constant of 75 N/m. The maximum potential energy stored in the spring is calculated to be 114.7 J, and the distance the system moves before stopping is determined to be 1.67 m. The key equations used include PE = 1/2kx² and d = 1/2kx²/mg.

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rocky4920
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Homework Statement


[/B]
A small cube, with a mass of 2 kg, slides along a frictionless horizontal surface at a constant speed of 20 m/s until it collides with, and sticks to, a large wooden 5 kg block. The large block is attached to the left end of a spring with a spring constant of 75 N/m as shown in the diagram.

upload_2018-1-11_20-38-51.png

a. What is the momentum of the cube before the collision?
b. What is the kinetic energy of the cube before the collision?
c. Find the speed of the combined cube and block system just after the collision.
d. Find the kinetic energy of the cube-block system just after the collision.
e. What is the maximum potential energy that can be stored in the spring due to this collision?
f. How far will the cube-block system move before it stops?

Homework Equations


PE= 1/2kx2
1/2kx2=mgd
d=1/2kx2/mg

The Attempt at a Solution



I have a question regarding a problem that I am trying to solve. I successfully have completed a,b,c,d and am having trouble with e and f. To find potential energy, 1/2kx2, I know the k value, but there is no x given in the problem. Is there another formula to use to solve? For f, I have the equation ready to go, but again I need to know x. For f, my equation is d=1/2kx2/mg.

Much thanks and appreciation.
 

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Forget f for the moment. Concentrate on e, and think how it relates to d.
 
Thank you.

KE = PE
1/2kx2 =1/2mv2
1/2(75)x2 =1/2(7)(5.71)2
x2=2(114.11)/75
x=1.749m

Solve for E
PE=1/2kx2
PE=1/2(75)(1.749)2
PE=114.7 J

Solve for F
d=1/2kx2/mg
d=114.7/7(9.8)
d=1.67 m

Is this the right track, or are there still errors I am making?
Thanks again!
 
rocky4920 said:
Solve for F
d=1/2kx2/mg
I do not understand what you are doing here. How does gravity come into it? This is all on a smooth horizontal surface.
You already found the answer as x.
 
Very true. I thank you again for your help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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