Physics question on Energy and Spring, any help appreciated

AI Thread Summary
The discussion revolves around a physics problem involving a puck compressing a spring on frictionless ice. For part a, the maximum compression of the spring is calculated using energy conservation principles, yielding a compression of approximately 2.535 meters. In part b, the introduction of a frictional force complicates the calculation, as the user struggles to determine the correct distance over which the friction acts. It is clarified that the distance Δd, where the puck compresses the spring, is the same as the compression distance x in part b, since friction acts from the moment the puck touches the spring. This understanding resolves the user's confusion regarding the relationship between the two parts of the problem.
DiaperBoy35
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Physics question on Energy and Spring, any help appreciated :D

Homework Statement



a. A 1.0 kg puck sliding at 15 m/s along some horizontal frictionless ice strikes and compresses a horizontal spring attached to one end of the ice rink. If the spring has a constant of 35 N/m, what is the maximum compression of the spring?

b. If in the above problem, the puck experiences a constant frictional force of 4.0 N opposing its motion beginning when it first strikes the spring, what would the maximum compression of the spring now be? (10 marks)


I am unsure about part b, as i am not sure what to use for the value of Δd as you will see underneath at my attempt

Homework Equations



-0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1)

- E(thermal) = Frictional force x Δd

The Attempt at a Solution



E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = 0.5kv^2 (1)
x = √((15^2)/35)
x = 2.535 m


[ (1) <-- denotes to initial (2)<-- denotes to final]

Part b
E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) + E(thermal) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = + 0.5kv^2 (1) + Ff Δ d

I am unsure of what to do from here

[ (1) <-- denotes to initial (2)<-- denotes to final]
 
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The amount by which the spring compresses, x, is the same as the distance over which work is done by friction, which you have called Δd.

Oh, and by the way, the elastic potential energy is given by (1/2)kx2. (It should be an x, not a v).
 


Oh yes, i meant kx :P
thank you for that

But why would Δd be the x value from part a?
 


DiaperBoy35 said:
Oh yes, i meant kx :P
thank you for that

But why would Δd be the x value from part a?

Δd is not the x value from part a. Δd is the x value from part b! :-p

The reason is because the frictional force starts up at the instant that the object touches the spring. So the distance over which the spring is compressed (before the thing comes to a stop) is the same as the distance over which the friction acts.
 


Ohhhhhh Thank youu so much!
I was really confused with that question lol
 
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