Amass of 0.75kg is attached to a spring and placed on a horizontal surface. the spring has a constant of 180 n/m, and the spring is compressed 0.3m past its natural length. if the mass is released from it's compressed position, what is the speed of the mass as it passes the natural length of the spring a. 2.1 m/s b. 0.83 m/s c. 3.8 m/s d.4.6 m/s

So we can solve this with the law of conservation of energy for the spring. The potential energy stored in the spring is Es=(1/2)*k*x^2 where k is the constant of the spring and x is the position. The kinetic energy of the spring is Ek=(1/2)*m*v^2 where m is the mass of the body attached to the spring and v is the speed of the body. When the spring is compressed, we store the energy into the spring and when the body is passing trough the natural length of the spring that potential energy is completely transformed into kinetic energy so it must be that Es=Ek.

Or (1/2)*k*x^2=(1/2)*m*v^2.

Now we solve for v: 

1/2 cancels out and we have: k*x^2=m*v^2.

v^2=(k/m)*x^2, now we take the second root of both left and the right side: 

v=√(k/m) * x, we input the numbers and get v=√(180/0.75)*0.3=15.5*0.3=4.65 m/s. So the correct answer is v= 4.6 m/s.

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