Question

A mass of 0.75 kg is attached to a spring and placed on a horizontal surface. The spring has a spring constant of 180 N/m, and the spring is compressed 0.3 m past its natural length. If the mass is released from this compressed position, what is the speed of the mass as it passes the natural length of the spring?

A. 4.6 m/s
B. 0.83 m/s
C. 3.8 m/s
D. 2.1 m/s

Answer 1

We can use the conservation of energy to solve this problem!
Initially, we're giving it spring potential energy. There's no kinetic energy yet because it's not releasing the energy in a form of movement! When we release the spring, all the potential energy is converted into kinetic energy!

\[\huge KE_0+PE_0=KE_f+PE_f\]
We already stated that there's no initial kinetic and no final potential, therefore
\[\huge PE_0=KE_f\]\[\huge \frac{ 1 }{ 2 }k(\Delta x)^2=\frac{ 1 }{ 2 }mv^2\]
Solve for v! :)

Also, notice here that there's no gravitational potential energy because we're not concerned here with changing it's position vertically. If we were to observe the block being shot from compressed spring up a ramp, then we would have to include the gravitational energy.

Answer 2

Can you help with filling in the variables? I'm really confused.

Answer 3

@CShrix

Answer 4

k is the spring constant
the change in x is the change in length of the spring due to compression/tension
m is the mass
v is the velocity (which you're supposed to find)

Using basic algebra techniques, you can isolate v to find a solution.