Collision of block with spring w/ friction

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Homework Help Overview

The problem involves a 4.20 kg block colliding with a horizontal spring with a spring constant of 270 N/m, compressing the spring by 4.00 cm. The coefficient of kinetic friction between the block and the surface is 0.250. Participants are exploring the work done by the spring, the energy dissipated by friction, and the speed of the block upon impact with the spring.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by the spring using the formula W = 0.5kx^2 but questions the correctness of their answer. They also calculate the work done by friction and express confusion regarding the speed of the block when it hits the spring.
  • Some participants question the assumptions made about the role of the spring at the moment of impact and the relationship between initial mechanical energy, work done by friction, and final mechanical energy.
  • Others suggest considering the total energy equation and the implications of the spring's potential energy.

Discussion Status

Participants are actively discussing the calculations and assumptions related to the problem. There is recognition of the potential for misunderstanding the role of the spring and friction in the energy equations. Some guidance has been offered regarding the total energy equation and the need to consider the work done by the spring.

Contextual Notes

There is mention of an online assignment indicating that the original poster's answer for part A is marked incorrect, leading to further exploration of the problem's assumptions and calculations. The discussion reflects uncertainty about the correctness of the provided answers and the implications of signs in calculations.

Selophane
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A moving 4.20 kg block collides with a horizontal spring whose spring constant is 270 N/m. The block compresses the spring a maximum distance of 4.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. What is the work done by the spring in bringing the block to rest?

When I first tried this I assumed it was simply:

W = 0.5kx^2 = (0.5)(270)(0.04)^2 = 0.216 J

However the answer is not matching with the correct one, not sure where i went wrong on such a simple problem... sorry if this is too easy, which I'm sure it is...

btw, there is a 2nd and 3rd part of the question

2nd part:

How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

I got this one right, simply: W = Ffs = (4.2)(9.81)(0.25)(0.04) = 0.412 J


finally, part 3:

What is the speed of the block when it hits the spring?

now i tried simply using (work of friction) = 0.5mv^2 but am not getting correct answer?

Thanks,
Chris
 
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Selophane said:
A moving 4.20 kg block collides with a horizontal spring whose spring constant is 270 N/m. The block compresses the spring a maximum distance of 4.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. What is the work done by the spring in bringing the block to rest?

When I first tried this I assumed it was simply:

W = 0.5kx^2 = (0.5)(270)(0.04)^2 = 0.216 J

However the answer is not matching with the correct one, not sure where i went wrong on such a simple problem... sorry if this is too easy, which I'm sure it is...

btw, there is a 2nd and 3rd part of the question

2nd part:

How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

I got this one right, simply: W = Ffs = (4.2)(9.81)(0.25)(0.04) = 0.412 J


finally, part 3:

What is the speed of the block when it hits the spring?

now i tried simply using (work of friction) = 0.5mv^2 but am not getting correct answer?

Thanks,
Chris
I get the same answer you do for part a.
For part c, what happened to the work done by the spring (its Potential Energy change) in your equation?
 
Thanks for the prompt response!

Well glad to see I'm not crazy and kept getting same answer for A, but it's saying it is not correct (this is an online assignment)

For C, i assumed the spring didn't matter since it was just as it hit the spring? Perhaps I'm misunderstanding the problem. And if i do need to use answer from A, i cannot get correct answer since it is still giving our answer as incorrect.

thanks again,
Chris
 
Selophane said:
Thanks for the prompt response!

Well glad to see I'm not crazy and kept getting same answer for A, but it's saying it is not correct (this is an online assignment)

For C, i assumed the spring didn't matter since it was just as it hit the spring? Perhaps I'm misunderstanding the problem. And if i do need to use answer from A, i cannot get correct answer since it is still giving our answer as incorrect.

thanks again,
Chris
The total energy equation is
Inititial mech. energy - work done by friction = final mech. energy

Since initially there is no PE, and in the end, no KE, then
KE_i -W_f = PE_f
1/2mv^2 - W_f = 1/2kx^2

You've already solved W_f and 1/2kx^2, so now solve for v.

BTW, the work done by a conservative force (like a spring force or gravity force) is equal to the change in PE. So your answer to part a is correct. What makes you think it is not, the book answer? Maybe you're a bit smarter than the book, yes?
 
ah, thanks for the clarification on C, however this is an online assignment and it does say 0.216 J is wrong for A, and well I can't do much if its saying its incorrect, I'll have to let the teacher know I guess, thanks again, much appreciated!


-Chris
 
If the spring compresses, why won't it uncompress, push the block away? Friction will do the job of stopping it.
 
Selophane said:
ah, thanks for the clarification on C, however this is an online assignment and it does say 0.216 J is wrong for A, and well I can't do much if its saying its incorrect, I'll have to let the teacher know I guess, thanks again, much appreciated!


-Chris
Ahh, try -.216J. That minus sign will do it. Sneaky!
 

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