A student has six textbooks, each with a thickness of 3.6 cm and a weight of 38 N. What is the minimum work the student would have to do to place all the books in a single vertical stack, starting with all the books on the surface of the table?
Just remember that the formula for Work is Force times Distance. You have the force which is 38 N. And N is Newtons if you didn't know what that was. And remember distance is always given in metres, so you have to convert cm to metres. There are 0.01 meters in a centimeter so you times 0.01 with 3.6 and you get 0.036 meters, and this is your distance. Then you times 38 by 0.036 to get your answer.
W = Fd
W = (38)(0.036)
W =
^ Can you do that?
Oh one thing you might be getting confused on is where it says --> weight of 38N. Weight and force are the same thing, so just watch out for that, it's something I got confused with too earlier.
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 gravitational potential energy.
The gravitational potential energy is given by the equation:
E = mgh
Where E is the gravitational potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height.
To calculate the mass of each book, we can use the formula:
Force (weight) = mass * acceleration due to gravity
Given that each book has a weight of 38 N, and the acceleration due to gravity is approximately 9.8 m/s^2, we can rearrange the formula to solve for mass.
mass = force / acceleration due to gravity
mass = 38 N / 9.8 m/s^2
mass ≈ 3.88 kg (rounded to two decimal places)
Now, let's find the height of the stack of books. Since there are six books and each book has a thickness of 3.6 cm, the total height is:
height = thickness * number of books
height = 3.6 cm * 6
height ≈ 21.6 cm = 0.216 m (rounded to three decimal places)
Now, we have all the values we need to calculate the gravitational potential energy:
E = mgh
E = 3.88 kg * 9.8 m/s^2 * 0.216 m
E ≈ 8.11 J (rounded to two decimal places)
Therefore, the minimum work the student would have to do to place all the books in a single vertical stack is approximately 8.11 Joules.