Posted by **Meghan** on Sunday, November 7, 2010 at 8:59pm.

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 block-table coefficient of kinetic friction?

- physics -
**Anonymous**, Monday, December 1, 2014 at 4:45pm
hhh

- physics -
**Ahmad**, Sunday, April 12, 2015 at 5:48pm
I want the answer. Where is it ?

- physics -
**Juan Niyebe heheh**, Monday, February 8, 2016 at 4:20am
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 Y-axis = 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

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