Work Done on a Book: 14 J or 42 J?

[mystry4]

Am I doing this correctly?
If a book that weighs 7 N is picked up off the floor and put on a shelf 2 m high, the work done on the book would be W = Fd (7N)(2m) = 14 J

If the same 7 N book is slid under the bookcase 3 m away, with a pushing force of 5 N and a frictional opposing force of 2 N, then the work done is

F = F(push) - f(friction) = 5 N - 2 N = 3 N then W = Fd (3N)(3m) = 9 J OR
should I be taking the 14 J from the first part and using
W = Fd = (14 J)(3m) = 42 J
Thank you for your help

 

[AKG]

Work = Force_applied * distance. Your first attempt at the second question is close, the second is way off. First of all, why would you be using the number from the previous question? In the first one, you're lifting a book, and in this one, you're pushing it. Second of all, 14 J is a measure of work, so how could you use that as your value for force? And why would 14 J x 3 m = 42 J? Look at the units.

 

[mystry4]

So, W = Force applied x distance would be 5N(the push) x 3 m = 8J.
The 2N of frictional force is one of those informational numbers and not
one used in the formula?

 

[AKG]

Yes, but 5 x 3 is not 8.

 

[mystry4]

Hmm... THAT is what I get for trying to do too many things at once! Of course, 5 x 3 = 15, geez...I need a vacation!

 

[HallsofIvy]

Notice that since the applied force (5 N) is greater than the friction force (2 N), the book will not just "slide" under the desk. There will be 2 N "net" force on it which will result in an acceleration. The book will wind up under the desk, not stopped but with a net speed (and so kinetic energy). That's where that additional 9 Joules goes.