Hooke's Law: Spring, Distance and Time

kgal

1. The problem statement, all variables and given/known data
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).


2. Relevant equations

F = -kx



3. 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

rock.freak667

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.

kgal

Thanks alot!

kgal

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...

gneill

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.