A 150 gram mass attached to a spring, which moves side to side on a fractionless

Question

A 150 gram mass attached to a spring, which moves side to side on a fractionless table. 64 joules of work is needed to squeeze the spring 40 cm.
1) the spring is squeezed fully ( the initial conditions needed here ) PE and KE ?
2) the spring is stretched fully. PE and KE ?
3) the spring is at its equilibrium length.PE and KE ?
4) the spring is squeezed fully ( the initial condition ) velocity?
5) the spring is stretched fully. velocity ?
6) the spring is at its equilibrium length. velocity ?
7) what is the spring constant K ?
8) when the spring is strecth only 20 cm, how much energy is stored in the spring? how much KE does the object have? what is the velocity of the mass?

Explanation / Answer

7) what is the spring constant K ? PE=(1/2)kx2 plug in x which is 40cm=0.40m and PE is 64J and solve for k. 8) when the spring is strecth only 20 cm, how much energy is stored in the spring? how much KE does the object have? what is the velocity of the mass? Plug in 0.20m into the previous equation for x and since you have k, now you can calculate the PE. Subtract that value from 64 to get the KE because Total Energy= KE + PE. And once you have KE you just set it equal to the KE equation i mentioned earlier and solve for v again. Good luck!

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