First post: work and energy problem

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Homework Help Overview

The problem involves a 20.0 kg block attached to a spring, where the block is pulled and released on a horizontal surface with friction. The objective is to determine the maximum kinetic energy attained by the block as it moves from the point of release to when the spring is unstretched.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles and the work-energy theorem but questions the validity of their calculations. Some participants raise concerns about the dimensional correctness of the equations used, while others suggest using the work-energy theorem as a clearer approach.

Discussion Status

Participants are actively discussing the validity of the original equations and exploring alternative methods to solve the problem. Some guidance has been offered regarding the application of the work-energy theorem, and there is acknowledgment of the need for clearer reasoning in the setup of the problem.

Contextual Notes

There are indications of confusion regarding the application of forces and energy principles, as well as the need for dimensional consistency in equations. The original poster expresses uncertainty about their final answer, which has been challenged by other participants.

Ianardo
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Homework Statement


A 20.0 kg block on a horizontal surface is attached to a horizontal spring of k = 2.0 kN/m. The block is pulled to the right so that the spring is extended 10.0 cm beyond its unstretched length, and the block is then released from rest. The frictional force between the sliding block and the surface has magnitude 80.0 N.

Homework Equations


What is the maximum KE attained by the block as it slides from release to the point where the spring is unstretched?

The Attempt at a Solution


KE is maximized when velocity is maximized. Therefore, the point is when the two forces are at equilibrium.
Set kx = Fd, d = (0.1-x)
2000(x) = 80(0.1-x)
x = 8/2080 (m)
= 0.003846 (m)

EPE (when fully stretched to 0.1 m) + Work done by friction = KE + EPE2
(.5)(2000)(0.1^2) - 80(0.1-0.003846) = .5(2000)(0.003846^2) + KE
Therefore, KE = 2.29 J, which makes sense.

I also did it the calculus way
W = the integral of force
F = ma, m =20 kg
a = [kx-f(0.1-x)]/20, integrating this yields velocity function, c =0
derive that function and set it to zero, yields the same answer.

But 2.29 J is not right. Why?
 
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Ianardo said:
Set kx = Fd
What are you doing here? Compare the units on either side of this equation!

In general, always check that the units on each side of your equations match and never forget to write out the units.
 
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Hi lanardo and welcome to PF.

Your starting equations are incorrect.
Ianardo said:
kx = Fd
This is dimensionally incorrect because kx has dimensions of [force] and Fd has dimensions of [Force×distance].
Ianardo said:
kx-f(0.1-x)
This is also incorrect for the same reason.
Do the problem by applying the work-energy theorem when the spring is at some distance x from its unstretched position and the mass is moving with speed v. Use symbols and put in the numbers at the very end so you can see what's going on.
 
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I know what you mean. But how would you solve it if you were to not equate the two forces?
 
Ok I see, thanks for the words!
 
Got it! Thanks for both of your help!

3.6 J is the final answer. I used the work-energy theorem and took the derivative AND set it to zero, find x. Then plug the x back into solve for KE.

Thank you for your advice. Merry Christmas!
mass%20on%20spring%20j.jpg
 

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