Potential energies and finding the distance

[Seung Lee]

Homework Statement

A horizontal spring with psring constant k = 15.19 N/m is compressed 23.11 cm fro mits equil. position.
A hockey puck with mass m = 170.0g is placed against the end of the spring. The friction coefficient is 0.02221.
How far does the hockey puck travel?

k = 15.19 N/m
x = 0.2311
uk = 0.02221
m = 0.17

Homework Equations

1/2mv^2 = 1/2kx^2 = friction force potential

The Attempt at a Solution

x = (kx^2)/ ( 2*0.02221*mg) = 10.9m

x should be equal to 23.11 cm...

I really don't understand why I keep getting a huge number when it shold actually be 23.11 cm. Also, I do understand tat 1/2mv^2 = 1/2kx^2 but why is any of those = to frictional force potential?

Thank you!

 

[NTW]

I like these problems, but they are often beyond my abilities. However, I'll try to help...

I believe that there are two phases. First, the spring decompresses, and its energy is partly used up to move the puck against friction. With the rest of energy that is left, the puck acquires a velocity v, and its kinetic energy is used up moving the puck away, at a constant deceleration, against the friction force...

 

[Seung Lee]

Isn't that why the equation is 1/2mv^2 = 1/2kx^2 = frictional potential? I mean another problem is that the answer keeps being 10.96m instead of actualyl beeing a small number

 

[ehild]

Isn't that why the equation is 1/2mv^2 = 1/2kx^2 = frictional potential?

The equation has no sense. You might have misread it. The second "equal" has to be "minus".

The initial elastic energy of the compressed spring is converted partly to the kinetic energy of the puck, and the other part is used as work against friction:

1/2 mv²=1/2 kx²-W(friction) where W(friction) = mgu_k and x=0.2311 cm is the compression of the spring.

ehild

 

[Seung Lee]

yeah, but you still can't solve for the distance can you? i mean you need the distance in W(friction) and you also don't ahve the velocity in 1/2mv^2, which means you have two unknowns in one equation.

 

[ehild]

First find the kinetic energy the puck acquires until it leaves the spring. During that time is travels x1=0.2311 cm. When the spring gets released the puck gets detached from the spring, and moves further, loosing energy because of friction. Calculate the distance x2 traveled till the puck stops. The total distance is x=x1+x2

 

[NTW]

First find the kinetic energy the puck acquires until it leaves the spring. During that time is travels x1=0.2311 cm. When the spring gets released the puck gets detached from the spring, and moves further, loosing energy because of friction. Calculate the distance x2 traveled till the puck stops. The total distance is x=x1+x2

And don't forget, Seung, that there's energy loss to friction also during the expansion of the spring.

 

[gneill]

The Attempt at a Solution

x = (kx^2)/ ( 2*0.02221*mg) = 10.9m

x should be equal to 23.11 cm...

Why do think that the total distance that the puck travels is 23.11 cm? Wasn't that the initial distance that the spring was compressed? I mean, it's possible that it could be true if the total energy lost to friction happens to be equal to the total energy initially stored in the spring. But the coefficient of friction looks to be too small for that to be true given that short distance (Compare the energy stored in the spring versus the energy lost to friction over that distance).

Your calculated result of around 11 m looks plausible to me.