physics
posted by Meghan .
You push a 2.0 kg block against a horizontal spring, compressing the spring by 15 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 75 cm from where you released it. The spring constant is 200 N/m. What is the blocktable coefficient of kinetic friction?

I want the answer. Where is it ?

Let K = Kinetic Energy
U = Potential Energy
Wf = Work done by friction
Ug = Gravitational Potential energy
Ue = Elastic Potential Energy
Law of conservation of energy states that:
K1 + U1 + Wf= K2 + U2
Given:
mass = 2 kg
x = 0.15 m
k= 200
d = 0.75 m
Solution:
K1 = 0 (because it is not moving)
U1 = Ug + Ue (Ug =0) (Ue = (1/2)(k)(x^2) )
K2 = 0
U2 = 0
0 + 0 + (1/2)(200)(0.15)^2 + Wf = 0 + 0
Wf = 2.25 J
Observe this. Since the block is moving to the right, the force of the friction is acting to the left.
Wf = (f)(d)cos(theta)
2.25 = friction (0.75)(cos180)
Force of the friction = 3 Joules
To find the coefficient of friction, find the normal force acting on the block
Summation of forces along Yaxis = 0
Normal force  Weight = 0
Normal force = (2 kg)(9.8 m/s^2)= 19.6 N
Friction = (coefficent) (Normal force)
3 Joules = (coefficient) (19.6 N)
Coefficient of Kinetic Friction = 0.15306
:D