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A student librarian lifts a 2.2 kg book from the floor to a height of 1.25 m. He carries the book 8.0 m to the stacks and places the book on a shelf that is.35 m above the floor. How much work does he do on the book?

I understand most of it, but I am unsure of what distance to use. Please help? ...?

Verified Answer :

The solution to your problem is as follows:


2.2Kg9.8m/s = 21.56N 

21.56N1.25m = 26.95J 


We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.

whitneytr12

Answer:

\(W_{tot}=W_{1}+W_{2}=mgh_{2}=7.55[J]\)

Explanation:

The work done on an object is the scalar product between force and displacement. So we can write the work:

\(W=F\cdot d=Fd\)

In our case F and d are parallels then we have a common product of their magnitudes.

Now, the total work will be the sum these two works:

  1. Work done when the student librarian lifts a 2.2 kg book from the floor to a height of h₁=1.25 m.
  2. Work done when he places the book on a shelf that is h₂=0.35 m above the floor.

Let's recall that the force in this problem is just the weight of the book. F=m*g

  • The first work will be: \(W_{1}=mgh_{1}\). F and h1 are parallels
  • The second work will be:\(W_{2}=-mg(h_{1}-h_{2})\) is negative because the vector force and the vector displacement are anti parallels.

Finally, the total work will be the sum of W₁ and W₂.

\(W_{tot}=W_{1}+W_{2}=mgh_{2}=2.2*9.81*0.35=7.55[J]\)

I hope it helps you!

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