a student has six textbooks, each has a thickness of 3.6cm and a weight of 28 N. What is the minimum work the student would have to do to place all of the books in a single vertical stack, starting with all the books on the surface of the table?
Since all of the books are on the table, the 1st book doesn't have to be
lifted:
d =d1 + d2 + d3 + d4 + d5 + d6,
d = 0 + 1*3.6 + 2*3.6 + 3*3.6 + 4*3.6
+ 5*3.6,
d = 0 + 3.6 + 7.2 + 10.8 + 14.4 + 18.
W = F(0 + 3.6 + 7.2 + 10.8 + 14.4 + 18)
W = F * 54cm,
W = 28 * 0.54m = 15.12 Joules.
To find the minimum work the student would have to do to place all the books in a single vertical stack, we need to calculate the total potential energy of the books.
The potential energy of an object is given by the equation PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
First, let's calculate the mass of one textbook. We can use the equation weight = mass x gravity, where weight is given as 28 N and gravity is approximately 9.8 m/s^2. Rearranging the equation, we have mass = weight / gravity.
mass = 28 N / 9.8 m/s^2
mass ≈ 2.86 kg
Now, let's calculate the height of the stack. Since there are 6 textbooks, the height of the stack will be the sum of the thicknesses of the textbooks.
height = 6 textbooks x 3.6 cm
height = 21.6 cm
However, we need to convert the height to meters, as the unit of gravity is in meters per second squared.
height = 21.6 cm / 100 cm/m
height = 0.216 m
Now, we can calculate the potential energy of the stack using the formula PE = mgh.
PE = (mass of one textbook) x (height) x (gravity)
PE ≈ 2.86 kg x 0.216 m x 9.8 m/s^2
PE ≈ 6.399 Joules
Therefore, the minimum work the student would have to do to place all the books in a single vertical stack is approximately 6.399 Joules.