Question

You stand inside of a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. What effect does dropping the sandbag out of the cart at the equilibrium position have on the maximum speed of the cart? Can someone please explain why it has no effect on the max speed of the cart?

*I know this is not math, but I'm not getting any responses in the physics forum and this problem is really bothering me a lot.

Answer 1

i'm very rusty on this stuff - what you have hear is Simple Harmonic Motion - sorry - cant remember

Answer 2

The math I was able to come up with suggests that is should matter. I'm thinking it has to do with the fact that the bag was dropped from equilibrium position maybe?

Answer 3

\[\omega=\sqrt{\frac{k}{m}}\]\[x(t)=A\cos(\omega t)\]\[v(t)=-A\omega\sin(\omega t)=-A\sqrt{\frac{k}{m}}\sin(\omega t)\]\[|v(t)_{max}|=A\sqrt{\frac{k}{m}}<A\sqrt{\frac{k}{m-m_{bag}}}\]this is wrong you say, so I wonder what I'm missing...

Answer 4

This is the explanation I was given, but it does not really make sense to me: Dropping the sandbag doesn't change the speed of the cart. Since energy doesn't leave the system at any time after the bag is dropped, the cart will always return to the equilibrium point with the same kinetic energy, and therefore the same maximum speed.

Answer 5

I just got the exact same formula as above by equating the kinetic and potential energies...

Answer 6

imagine two objects side by side attached only by a piece of tape or something. The energy of the system is given by \[ME = KE + PE = \frac{1}{2}mv^2+\frac{1}{2}kx^2\]If you let go of the Sandbag when you are moving quickly past the equilibrium point, the sandbag hurtles along side you and carries away lots of kinetic energy. If you let go of it when the spring is at maximum stretch, it does not remove energy from your system. ...

Answer 7

I understand that Broken Fixer, but my problem is why does it not lose max velocity when it loses kinetic energy.

Answer 8

it loses kinetic energy due to mass. Since less mass is there less KE is needed to keep the max velocity. Imagine two pendulums swinging side by side, if you take one away the other is unaffected...

Answer 9

At the equilibrium point, you let go of the sandbag. You are now a system traveling at speed v_max, with new mass and with new kinetic energy KE_max. Your PE is PE = 0 currently.

Your explanation works fine - you go shooting off (less far), then come back to your same max speed on the way back...