Ball attached to String (Potential Energy)

[goluigi2196]

Homework Statement

A 2.40 kg ball is attached to a ceiling by a 2.00 m long string. The height of the room is 3 m. What is the gravitational potential energy of the ball relative to:

a) the ceiling?

b) the floor?

c) a point at the same elevation as the ball?

Variables
P for potential energy
m for mass in kg
g for gravity
h for height

Homework Equations

P=mgh

The Attempt at a Solution

For b), I found the answer which was 23.52.

For a), though, I got 70.56 because P=(2.4)(9.8)(3)=70.56. The 3 is from the fact that the ceiling is 3m off the ground and a) is asking about the ceiling. It says that I'm wrong

For c), I don't have any idea of what they're talking about. Is it the same thing as b)?

 

[zhermes]

Energy is always relative. Potential energy is defined as the potential energy of one point, vs. the potential energy at another point. The equation, more precisely, should be written U = mg \Delta h for some different in height \Delta h \equiv h - h_0.

Usually the 'reference' point (h_0) is taken to be "zero height" (h_0 = 0), and that is often either sea-level, or ground-level, or floor-level, etc.

For a), though, I got 70.56 because P=(2.4)(9.8)(3)=70.56. The 3 is from the fact that the ceiling is 3m off the ground and a) is asking about the ceiling. It says that I'm wrong

While the ceiling is 3m off the ground (h_0 = 3), it is only 1m away from the ball (the ball is what you're finding the potential energy of). h = 3m - 2m = 1m

For c), I don't have any idea of what they're talking about. Is it the same thing as b)?

If they're asking for the potential of the ball with respect to something at the same height, what is the difference in height \Delta h?

 

[goluigi2196]

ok, so I did 2.4(9.8)(1)=23.52 for a). i still get it wrong...

and for c), do i just do 2.4(9.8)(2) because it's the inverse?

 

[zhermes]

ok, so I did 2.4(9.8)(1)=23.52 for a). i still get it wrong...

Sorry, forgot to highlight a key point. In this case, the ball is lower than the reference point. I.e. \Delta h = h - h_0 = 2m - 3m

and for c), do i just do 2.4(9.8)(2) because it's the inverse?

Its not the inverse problem. Its asking what is the potential difference between something at h = 2m, and a reference point at h_0 = 2m

 

[Vespa71]

..attached to a ceiling by a 2.00 m long string

How far is it between ball and ceiling? Ball and floor?

 

[goluigi2196]

so will a) be negative because 2-3=-1? therefore, will the answer be -23.52?

and will c) be zero because 2-2=0 and 2.4(9.8)(0)=0?

 

[goluigi2196]

@vespa71

the problem said the ceiling to the floor was 3m

 

[Vespa71]

a) will be negative becaus there's a 2! meter negative drop from the ball to the ceiling. c) is zero as there's no drop. Well done.

 

[Vespa71]

I recommend to make a simple drawing to visualize the problem. Best of luck

 

[goluigi2196]

well ok, i did 2.4(9.8)(-1). that gives me -23.52. it still tells me I'm wrong

 

[goluigi2196]

oh and thanks for c). i got it right.

 

[Vespa71]

If you have a -2m drop from ball to ceiling, and a 1m drop from ball to floor, and a 0m drop from ball to somthing on the same level, I think it will solve.