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A physics textbook is launched up a rough incline with a kinetic energy of $X$ joules. When the book comes momentarily to rest near the top of the incline, it has only gained ($X-20$) joules of gravitational potential energy. About how much kinetic energy will it have when it returns to the launch point? (Hint: Remember friction and conservation of energy).

Attempt: So when the textbook goes up and reaches the point where it is at rest its GPE is $X$ joules. When it goes down and reaches the starting point its kinetic energy should by $X$ joules, I guess. Thats what I am not sure about. Does the friction act with the same force when the textbook goes down?

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Hi Dostre - this is an example of the sort of homework question that is forbidden in the FAQ. The general rule with these is that if you edit it to make it a conceptual question, we'll happily reopen it. It might help you to look at some of the good and bad examples of how to ask homework questions in our homework policy on meta. (Scroll down toward the bottom of the page for the examples.) – David Z May 11 '12 at 6:23
Something like that? I removed numbers and added variables which makes it general:). – Koba May 11 '12 at 6:34
Hey guys isn't now middle of the night there in the States? Get a sleep and read it tomorrow. – Pygmalion May 11 '12 at 6:39
Sure, Dostre, that helps. What makes the difference is not really the substitution of variables for numbers, but rather the text you've added below the problem in which you explain the core concept that is confusing you: namely whether friction acts the same way in both directions. That's what the question should focus on. (I'm going to fix up your title to match.) – David Z May 11 '12 at 7:54
@DavidZaslavsky Changing the question is fine and I approve that in this case, but generally the consequence is that the answer no longer match the question. Which will result in this case that someone like Ron will come along and criticize the answer, possibly downvoting it etc. – Pygmalion May 11 '12 at 8:06
up vote 2 down vote accepted

According to rules of the forum, we are not allowed to solve the problems, but discuss physical principles that lead to the solution. Here is my guidance.

The law of conservation of energy states

$$W_\text{NC} = E_\text{final} - E_\text{init}$$

where $W_\text{NC} = \vec{F}_\text{NC} \cdot \vec{s}$ is work of non-conservative force(s), while $E$ is mechanical energy (kinetic + potential). As you see, the change in mechanical energy equals the work of non-conservative forces, in your case the work of force of friction.

Now, what you have to do is estimate, what is the amount of work of non-conservative forces when textbook is going up and when textbook is going down. What is the magnitude and the direction of $\vec{F}_\text{fr}$ and displacement vector $\vec{s}$ in each case? Don't forget that $\vec{F} \cdot \vec{s}$ is scalar product!

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I see. So, When the textbook goes up the work of non-conservative forces (friction) is ($-20$) joules. Thus, when the object goes down its Potential energy is $180$ joules. When it gets back to the starting point its kinetic energy will be $160$ joules since the friction was ($-20$) joules. I had this idea in mind. I just did not know whether the friction was the same when the textbook was going up and when it was going down. Thanks. – Koba May 11 '12 at 6:14
Yes, this is correct. The force of friction has constant magnitude and the magnitude of the displacement is the same in both directions. Since vector of friction and vector of displacement are in opposite directions (first displacement up and friction down, later displacement down and friction up), you get negative work (scalar product!), which amounts to -20 J in both directions. – Pygmalion May 11 '12 at 6:19
Thanks sir. You are very helpful. A bit more rep and I will upvote your answers. – Koba May 11 '12 at 6:21
You can have materials where friction is direction dependent (ex sharkskin, barbed points); but materials like that are the exception to the rule. Unless explicitly stated otherwise, you should assume there's no directional asymmetry. – Dan Neely May 11 '12 at 18:12

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