Math / Calculus

Question10. (II) In a certain library the first shelf is 15.0 cm off the ground, and the remaining four shelves are each spaced 38.0 cm above the previous one. If the average book has a mass of 1.25 kg with a height of 22.0 cm , and an average shelf holds 28 books (standing vertically), how much work is required to fill all the shelves, assuming the books are all laying flat on the floor to start?

Studdy Solution

STEP 1

1. The first shelf is 15.0 cm above the ground.
2. Each subsequent shelf is 38.0 cm above the previous one.
3. There are a total of 5 shelves.
4. Each book has a mass of 1.25 kg.
5. Each book has a height of 22.0 cm.
6. Each shelf holds 28 books.
7. The books start on the floor.

STEP 2

1. Calculate the height of each shelf.
2. Determine the gravitational potential energy change for moving one book to each shelf.
3. Calculate the total work required for all books on each shelf.
4. Sum the work for all shelves.

STEP 3

Calculate the height of each shelf:
- First shelf: $h_1 = 15.0 \, \text{cm}$ - Second shelf: $h_2 = h_1 + 38.0 \, \text{cm} = 53.0 \, \text{cm}$ - Third shelf: $h_3 = h_2 + 38.0 \, \text{cm} = 91.0 \, \text{cm}$ - Fourth shelf: $h_4 = h_3 + 38.0 \, \text{cm} = 129.0 \, \text{cm}$ - Fifth shelf: $h_5 = h_4 + 38.0 \, \text{cm} = 167.0 \, \text{cm}$
Convert heights to meters:
- $h_1 = 0.15 \, \text{m}$ - $h_2 = 0.53 \, \text{m}$ - $h_3 = 0.91 \, \text{m}$ - $h_4 = 1.29 \, \text{m}$ - $h_5 = 1.67 \, \text{m}$

STEP 4

Determine the gravitational potential energy change for moving one book to each shelf using the formula:
$\Delta U = m \cdot g \cdot h$
where $m = 1.25 \, \text{kg}$ and $g = 9.81 \, \text{m/s}^2$ .
Calculate for each shelf:
- First shelf: $\Delta U_1 = 1.25 \times 9.81 \times 0.15$ - Second shelf: $\Delta U_2 = 1.25 \times 9.81 \times 0.53$ - Third shelf: $\Delta U_3 = 1.25 \times 9.81 \times 0.91$ - Fourth shelf: $\Delta U_4 = 1.25 \times 9.81 \times 1.29$ - Fifth shelf: $\Delta U_5 = 1.25 \times 9.81 \times 1.67$

STEP 5

Calculate the total work required for all books on each shelf:
Each shelf holds 28 books.
- First shelf: $W_1 = 28 \times \Delta U_1$ - Second shelf: $W_2 = 28 \times \Delta U_2$ - Third shelf: $W_3 = 28 \times \Delta U_3$ - Fourth shelf: $W_4 = 28 \times \Delta U_4$ - Fifth shelf: $W_5 = 28 \times \Delta U_5$

STEP 6

Sum the work for all shelves to find the total work required:
$W_{\text{total}} = W_1 + W_2 + W_3 + W_4 + W_5$
Calculate each:
- $\Delta U_1 = 1.25 \times 9.81 \times 0.15 = 1.84 \, \text{J}$ - $\Delta U_2 = 1.25 \times 9.81 \times 0.53 = 6.49 \, \text{J}$ - $\Delta U_3 = 1.25 \times 9.81 \times 0.91 = 11.16 \, \text{J}$ - $\Delta U_4 = 1.25 \times 9.81 \times 1.29 = 15.81 \, \text{J}$ - $\Delta U_5 = 1.25 \times 9.81 \times 1.67 = 20.46 \, \text{J}$
Calculate work for each shelf:
- $W_1 = 28 \times 1.84 = 51.52 \, \text{J}$ - $W_2 = 28 \times 6.49 = 181.72 \, \text{J}$ - $W_3 = 28 \times 11.16 = 312.48 \, \text{J}$ - $W_4 = 28 \times 15.81 = 442.68 \, \text{J}$ - $W_5 = 28 \times 20.46 = 573.04 \, \text{J}$
Total work:
$W_{\text{total}} = 51.52 + 181.72 + 312.48 + 442.68 + 573.04 = 1561.44 \, \text{J}$
The total work required to fill all the shelves is:
$\boxed{1561.44 \, \text{J}}$

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