Question

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? ...?

Answer 1

To calculate the work done on the book, the student librarian must use the formula W = mgh for each segment of lifting the book vertically. The total work done is the sum of the work for lifting the book to 1.25 m and then to 0.35 m. The horizontal distance carried does not contribute to work against gravity.

Step-by-step explanation:

The work done on the book by the student librarian can be calculated by considering the work done against gravity to lift the book to different heights. The distance carried horizontally does not involve work against gravity under the assumption that there is no horizontal force like friction opposing the motion. The work done to lift the book from the floor to 1.25 m can be calculated using the formula W = mgh, where m is the mass of the book, g is the acceleration due to gravity (approximately 9.8 m/s2), and h is the height.

The initial lift is W = 2.2 kg × 9.8 m/s2 × 1.25 m. When placing the book from the carried position to a shelf at 0.35 m height, the work done is W = 2.2 kg × 9.8 m/s2 × 0.35 m since the vertical displacement is from 1.25 m to 0.35 m, not the full 1.25 m again.

The total work done on the book is the sum of these two amounts of work. Please note that the gravitational constant (g) could be given or assumed in this context. The exact value used may vary slightly depending on the context of the problem.

Answer 2

`W_(tot)=W_(1)+W_(2)=mgh_(2)=7.55[J]`

Step-by-step 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!