How to calculate work required to move an object

[ACasey]

Homework Statement

Eight books, each 4.3 cm thick with a mass of 1.7 kg, lie flat on a table. How much work is required to stack them one on top of another?

Homework Equations

w= Fdcos(\theta

The Attempt at a Solution

For the first book, there is no work required. The displacement was zero therefore no work was required. For the second book, the d=4.31 since the book must be lifted up, but I am not sure how to calculate the force and figure the angle. Don't I need to know where the books are to begin with?

 

[Irid]

Your angle is \theta = \pi/2 here, because the force is vertical. For the force, use gravitational pull: F=mg, where g=9.8 \unitfrac[m][s^2] is the free acceleration.

 

[ACasey]

If I understand this right, for my second book, I would have 112 J. I took \pi//2 = 1.57 then multiplied that by 9.8*1.7 kg and got 26.16, then multiply that by the 4.3 cm and got 112 J for my final answer. then for the third book instead of multiplying it by 4.3 I would multiply it by 8.6 cm since I will have two books under it already.

 

[Irid]

I get a different result. Please write down your final expression in symbolic form.

 

[ACasey]

I apologize, but I am not sure what symbolic form is. Can you please explain further?
Thank you

 

[Feldoh]

Your force and displacement are in the same direction so W = F*d

 

[Irid]

ACasey, symbolic form is a fancy name for writing as much as possible in letters. That way, initial energy is

E_1 = 8mgh/2 = 4mgh

if h=4.3cm and it labels the height of center of mass. The final energy is

E_2 = 8mg \frac{8h}{2} = 32mgh

The difference is then

E_2-E_1 = 32mgh - 4mgh = 28mgh

This is the answer in symbolic form. Plugging in the numbers we obtain the answer in numeric form.

 

[ACasey]

I guess I was unaware of how confused I am with this subject. How is energy coming into this problem? I thought the formula that I would use is force which is F= mass*gravity then multiply that by the displacement which in this case would be 4.3 c.m. for each book. Then I multiply that by the \pi2 and that would be the solution in joule? Where am I getting lost?

 

[Irid]

If you multiply force times displacement, the result is energy. The angle is not required here, so forget about it. You can follow my solution, or if you insist on your way, then the displacement for the first book is h_1 = 0, for the second book h_2 = h, for the third h_3 = 2h and so on, then you add all them and the result obtains. Don't forget to convert everything to SI units if you want to get joules in the end.

 

[ACasey]

I'm sorry, I didn't mean to offend you. I would like to do it the correct way, but I am confused on the solution. I will get it figured out.
Thank you for your help.

 

[Irid]

You didn't offend me. There are several ways to calculate this result, doesn't matter too much which is used.
Basically this is just conservation of energy problem.

 

[Natalie<3]

What about ...; A force of 30N does 180J of work across the floor . If the same
amount of work is done over a distance of 6m, how much force is doing work ?