Potential Energy of Elastic Band

[ppkjref]

Elastic Potential Energy of Elastic Band*

Homework Statement

A stretched elastic band of mass 0.55 g is released so that its initial velocity is horizontal, and its initial position is 95 cm above the floor. What was the elastic potential energy stored in the stretched band if, when it lands, it has a horizontal displacement of 3.7 m from the initial position under negligible air resistance?

Homework Equations

vf^2 = vi^2 + 2ad
Ee = Ek - Eg

The Attempt at a Solution

vf^2 = vi^2 + 2ad
0 = vi^2 + 2ad
2(9.8)(0.95) = vi^2
4.32 m/s = vi

Ee = Ek - Eg
Ee = 0.5mv^2 - mgh
Ee = (0.5)(0.55)(4.32)^2 – (0.55)(9.8)(3.7)
Ee =
I get a negative number?

 

[berkeman]

Elastic Potential Energy of Elastic Band*

Homework Statement

A stretched elastic band of mass 0.55 g is released so that its initial velocity is horizontal, and its initial position is 95 cm above the floor. What was the elastic potential energy stored in the stretched band if, when it lands, it has a horizontal displacement of 3.7 m from the initial position under negligible air resistance?

Homework Equations

vf^2 = vi^2 + 2ad
Ee = Ek - Eg

The Attempt at a Solution

vf^2 = vi^2 + 2ad
0 = vi^2 + 2ad
2(9.8)(0.95) = vi^2
4.32 m/s = vi

Ee = Ek - Eg
Ee = 0.5mv^2 - mgh
Ee = (0.5)(0.55)(4.32)^2 – (0.55)(9.8)(3.7)
Ee =
I get a negative number?

I don't think you can set vf = 0. It could have been still moving horizontally when it hit the floor.

Instead, relate the initial KE stored to the initial horizontal velocity as one equation. Then relate the time of flight to the distance traveled and the horizontal velocity (which won't change over time). And how long does it take something to drop vertically from that height...?

 

[ppkjref]

I'm not sure what you mean. Initially wouldn't there only be elastic potential stored? KE is 1/2mv^2 and the horizontal velocity is d/t? I don't know what time is

 

[berkeman]

I'm not sure what you mean. Initially wouldn't there only be elastic potential stored? KE is 1/2mv^2 and the horizontal velocity is d/t? I don't know what time is

Yes, initially at time t=0-, all energy is stored as PE. But as the band is snapped and starts flying at t=0+, all that PE was converted into KE of horizontal motion. The horizontal velocity remains constant (ignoring air resistance), and the vertical motion follows the usual kinematic equations of motion for constant acceleration (gravity).

The "time" is the time between release and hitting the floor. Try writing the equations out that I alluded to in my previous post, and see where they lead you. I'll try to check back in a few hours...

 

[ppkjref]

Ok I got 0.019 J as my final answer. Thanks for the help