physics lo1 hooke's law horizontal spring with weight problem

1. Horizontal Spring and Weight on Frictionless Surface Involving Hooke’s Law
Consider the following image:
A horizontal spring with a weight of 300 grams attached to one end resting on a frictionless
surface is at equilibrium with the spring length being 13.0 cm long. Then, someone pushes
the weight towards the left, compressing the spring until the spring’s length is only 10.0 cm.
Then the spring is released and the weight moves to the right. If the spring constant is
120N/m, what is the potential energy of the spring when compressed to 10 cm? What is the
horizontal speed of the weight after it has moved 3.0 cm to the right upon being released?
How about 4 cm?

2.
ANSWER:
Potential energy of spring when compressed to 10 cm = 0.054 J
Speed at 3 cm = 0.60 m/s
Speed at 4 cm = 0.57 m/s
EXPLANATION:
When dealing with problems with springs, it is necessary to have a good understanding of
Hooke’s Law and applying it. Hooke’s Law states that the force needed to extend or compress
a spring by a distance is proportional to that distance and is shown by F = ­kx. F is the force
used or needed, k is the spring constant, which is an inherent factor of any spring, and is
related to a spring’s stiffness. While x is the distance which the spring is compressed or
extended. Lastly, the negative sign simply denotes that the force that the spring exerts is
always in the opposite direction from the force acting on it, for example, if you’re compressing
a spring by pushing left, it will apply a force towards the right.
By applying energy conservation, the energy in the instance where the spring is compressed
is Potential Energy + Kinetic Energy, which in that case would be (½)kx2
+0, since the mass
isn’t moving. Plugging in the spring constant and distance compressed (13 cm ­ 10 cm = 3
cm) to that equation will give you (½)(120 N/m)(0.030 m)2
+0 = 0.054 N*m which is 0.054 J.
When the mass has moved 3 cm to the right, it is back to the original position where the
spring is uncompressed and at it’s natural length. Since there is no longer any potential
energy in the spring, all the energy must be kinetic energy and with that information, the
equation for the total energy at that time would be 0+(½)mv2
. From the previous paragraph
above, we know that the total energy in the situation is 0.054 J. Therefore, 0+(½)mv2
= 0.054
J. Rearranging the equation to solve for velocity, we get v = (((0.054 J)*2)/m)0.5
= (((0.054
J)*2)/(0.3 kg))0.5
= 0.60 m/s
We know that the total energy in this situation is 0.054J. If the weight has moved 4 cm to the
right from where it has been released, it is 1 cm to the right of equilibrium for the spring and
so we can solve for this equation: 0.054J = (½)kx2
+(½)mv2
Isolating v, we get the equation: v
= (((0.054 J)­(½)kx2
)/0.5m)0.5
= (((0.054 J)­(½)(120 N/m)(0.01 m)2
)/((0.5)(0.3))0.5
= 0.57 m/s