# Homework Problem about the Work-Energy Theorem

1. Nov 22, 2007

### aquamarine08

[SOLVED] Homework Problem about the Work-Energy Theorem

A puck on a shuffleboard decreases in speed from 1.0 m/s to .20 m/s while traveling 4.0 m. What is the coeffecient of kinetic friction between the puck and the board?

I think that the equations we have to use to solve this would be....

w= $$\frac{1}{2}$$m$$V_{1}^{2}$$-$$\frac{1}{2}$$m$$V_{o}^{2}$$

w=Fd

$$F_{f}$$=$$\mu$$$$F_{N}$$

w = $$\frac{1}{2}$$k$$d^{2}$$

k= $$\frac{1}{2}$$m$$V^{2}$$

First : I was thinking about using the equation w= $$\frac{1}{2}$$m$$V_{1}^{2}$$-$$\frac{1}{2}$$m$$V_{o}^{2}$$ to solve for work, then substitute the work value in for w=Fd then somehow put it in the equation $$F_{f}$$=$$\mu$$$$F_{N}$$ . Then I got extremely confused...and I realized that I don't have the mass of the puck to put into the first equation.

Second try: So, I decided to use w = $$\frac{1}{2}$$k$$d^{2}$$ but in order to find out the k (spring constant) value, I need to use the equation k= $$\frac{1}{2}$$m$$V^{2}$$...which again, I dont have the mass for.

Third try...I was going to use the equation F=kd to figure out F and then substitute it in for w in w=Fd and then solve for m in w= $$\frac{1}{2}$$m$$V_{1}^{2}$$-$$\frac{1}{2}$$m$$V_{o}^{2}$$
. But I don't have the k again!!

And now, I'm out of ideas...please help me...thank you! :) I hope this question makes sense....and my tries...if not, let me know and I'll try to fix them up! Thanks again!

2. Nov 22, 2007

### rl.bhat

Find the retardation of the body by using kinematic equatuon. Retarding force = m*a and normal reaction = mg. Plug it in the equation to get coefficient kinetic friction

3. Nov 22, 2007

### photonsquared

Try this

Force of friction is given by:

$$F_{f}$$ = $$\mu$$ N [1]

Solving for $$\mu$$ we have

$$\mu$$ = $$\frac{F_{f}}{N}$$ [2]

The normal force N is given by:

N = mg [3]

Substituting [3] into [2] we have:

$$\mu$$ = $$\frac{F_{f}}{mg}$$ [4]

The force of friction is also defined by:

$$F_{f}$$ = ma [5]

where a is the deceleration and m is the mass of the puck. We also know given a distance and a constant acceleration (or deceleration in this case) that the acceleration is equal to the following (see: http://library.thinkquest.org/10796/ch3/ch3.htm ):

a = $$\frac{(V_{f}^{2} - V_{i}^{2})}{2d}$$ [6]

Substituting [6] into [5] we have:

$$F_{f}$$ = $$\frac{m(V_{f}^{2} - V_{i}^{2})}{2d}$$ [7]

and finally substituting [7] into [4] we have:

$$\mu$$ = $$\frac{(V_{f}^{2} - V_{i}^{2})}{2dg}$$

4. Nov 23, 2007

### aquamarine08

photonsquared, I used your method to solve the problem....it made sense to me and this is the answer that I got $$\mu_{k}$$= .0122. Is there anyway you could tell me if this is correct? Thank you very much for your help!

5. Nov 23, 2007

### photonsquared

That is the same value I calculate. The reason you don't need the mass of the puck is because the masses cancel out when you substitute eqn 7 into 4. Hope this helps.

6. Nov 23, 2007

### aquamarine08

Ok thank you very much photonsquared! :) Thanks rl.bhat for your response!