Force of Spring and Potential Energy of Spring Pushing a Box

[RadiantL]

Homework Statement

A horizontal spring with spring constant 97.9 N/m is compressed 18.2 cm and used to launch a 2.96 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.156. How far does the box slide across the rough surface before stopping?

Homework Equations

Usp = spring potential = 1/2 (k)Δx^2
F = -kΔx
Work = FΔx

The Attempt at a Solution

Hi, so I'm not really looking for the answer... I got the answer oddly, there is just this little weird thing I'm dealing with.

Ok so I found the potential energy from the spring using the equation above and got 1.6 J now because the block is sent flying the potential from the spring is converted to kinetic, which should be the same amount I believe...

Now I was wondering why when I find the Force of the spring which comes to be 17.8178 N and multiply by the compression which is 0.182m, I don't get the same number as the potential energy? The work is done on... the block which has no kinetic energy, being at rest soooo the work should be equal to the change in kinetic energy which should be +1.6 J, but when I multiply 17.8178 with the compression distance 0.182m I get 3.24 J which is about 2x 1.6 which is a coincidence? or I am missing something.

Anyway yeah this seems like something that should be fine but I'm dazzled by the different numbers I'm getting, any clarification would be awesome

 

[Infinitum]

Hi RadiantL!

The work done by the spring varies with x, so you cannot use the simple work formula, W=F.d, as this gives you work for a constant force. You would have to integrate to find the actual work, whose magnitude, in fact would come out equal to the potential energy of the spring. That's why you do not get an exact answer from F.d, but instead get double the value, because force is linearly varying, which you can verify from the integral that would give you a 1/2 term.

Hope I didn't confuse you further.

 

[azizlwl]

The force in not constant, proportional to distant.
If you use integral, the ∫f(x)dx=1/2kx²

 

[RadiantL]

Ah I understand now! thank you very much, both of your replies cleared things up pretty nicely :P It is very much appreciated!

 

[asteriatic]

!

Hi there,

It seems like you have the right idea, but you may be missing a key concept. When the spring is compressed, it stores potential energy in the form of elastic potential energy. This energy is not equal to the force times the compression distance, but rather it is equal to the work done by the force in compressing the spring. In other words, the potential energy of the spring is equal to the work done on the spring, not the work done by the spring.

In this case, the work done on the spring is equal to the force of the spring times the compression distance, or 17.8178 N * 0.182 m = 3.24 J. This energy is then stored in the spring as potential energy, which is later converted to kinetic energy when the spring is released.

When the box is launched, the potential energy of the spring is converted to kinetic energy, as you mentioned. This means that the kinetic energy of the box is equal to the potential energy of the spring, or 1.6 J. However, when the box travels across the rough surface, it loses some of this kinetic energy due to friction. This is where the coefficient of kinetic friction comes into play.

The work done by friction is equal to the force of friction times the distance traveled. In this case, the force of friction is equal to the coefficient of kinetic friction (0.156) times the normal force (mg). The normal force is equal to the weight of the box, or mg = 2.96 kg * 9.8 m/s^2 = 29.008 N. Therefore, the work done by friction is 0.156 * 29.008 N * d = 4.524 d J.

Since the box stops moving when all of its kinetic energy is converted to work done by friction, we can set the kinetic energy equal to the work done by friction and solve for the distance d:

1.6 J = 4.524 d J
d = 0.353 m

Therefore, the box slides approximately 0.353 m across the rough surface before stopping.

I hope this helps clarify the different numbers you were getting. Keep up the good work!