Hooke's Law: Spring, Distance and Time

AI Thread Summary
The discussion revolves around a physics problem involving a mass moving towards a spring, where Hooke's Law is applied to determine the maximum compression of the spring and the total time for the mass to return to its starting point. The correct approach for part (a) involves using energy conservation principles, specifically equating the change in kinetic energy to the work done by the spring. For part (b), the motion can be analyzed using the equation for harmonic motion, considering the time taken for the mass to travel to the spring, compress it, and return. Participants emphasize the need to break the problem into segments based on the mass's interaction with the spring. The correct solutions are a maximum compression of 0.0656 m and a total elapsed time of 0.245 seconds.
kgal
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Homework Statement


A mass M = 0.454 kg moves with an initial speed v = 2.88 m/s on a level frictionless air track. The mass is initially a distance D = 0.250 m away from a spring with k = 876 N/m, which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance d, before reversing direction. after bouncing off the spring, the mass travels with the same speed v, but in the opposite direction.
a. Determine the maximum distance that the spring is compressed (0.0656 m).
b. Find the total elapsed time until the mass returns to its starting point (0.245 s).


Homework Equations



F = -kx



The Attempt at a Solution



a. Tried to use energy conservation equation but couldn't get the correct answer.
1/2mv0^2 + 1/2kd^2 + 1/2m1v^2 = 0.
got d = √(2mv^4/k) which got me an incorrect answer.

b. have no idea where to begin
 
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kgal said:

The Attempt at a Solution



a. Tried to use energy conservation equation but couldn't get the correct answer.
1/2mv0^2 + 1/2kd^2 + 1/2m1v^2 = 0.
got d = √(2mv^4/k) which got me an incorrect answer.

The line in red is incorrect. The change in KE = work done by the spring. So you should have

1/2m(v2-v20 = 1/2 kd2

And you know that v=0. This should get you the correct value for d.
 
Thanks alot!
 
for section b,
I was thinking of using the equation x(t) = Acos(sqrt(k/m)t) where x(t) = 0.25 m, k from the answer in section a and the given mass, but can't get anywhere when trying to solve for t...
 
kgal said:
for section b,
I was thinking of using the equation x(t) = Acos(sqrt(k/m)t) where x(t) = 0.25 m, k from the answer in section a and the given mass, but can't get anywhere when trying to solve for t...

The mass starts out 0.250m away from the spring... it's not even in contact with it. It first travels that distance at its initial constant velocity before coming in contact with the spring. Then it compresses the spring by the distance found in part (a) and is subsequently launched back along its initial trajectory.

You'll have to divide the problem into separate parts corresponding to when it is in contact with and not in contact with the spring. When it's in contact with the spring you can use properties of a mass-spring system that you should know.
 
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