A spring with a force constant of 200 N/m is compressed 20cm. How fast will a 3kg box placed on the top of the compressed spring be moving when the spring returns to its resting position?

Answer 1

Given

K = 200 N/m

Xa = 0.20 m

M = 3kg

Find

Velocity of the box when the spring is in resting position

Solution

Let compressed position be point A

Let resting position be point B

Apply conservation of Potential energy

PEa = m*g*ha + (1/2)*K*Xa^2

PEa = 0 + (1/2)*200*(.20)^2 = 4 J

Please note we set the point of measurement for gravitational potential energy at A. Therefore, mgha = 0

Now kinetic energy, at point A when the spring is compressed the velocity is 0

KEa = (1/2)*M*Va^2 = 0

Now, we find the Potential energy at point b

PEb = mghb + (1/2)*K*Xb^2

Spring Potential energy is zero

Therefore,

PEb = mghb = 3*9.81*0.2 = 5.886 J

Now, kinetic energy at point B

KEb = (1/2)*M*Vb^2 = (1/2)*3*Vb^2

Conservation of energy equation:

PEa+KEa = PEb+KEb

Sub in values discussed above

4 J + 0 = 5.886 J + (1/2)*3*Vb^2

(4-5.886) * 2/3 = Vb^2

-1.257 = Vb^2

Vb = Imaginary number

This means the box is too heavy and that you need a stiffer spring to propel it upward to the springs resting position. Picking a stiffer spring would allow you to overcome the gravitational potential energy of the box.