Conservation of Energy problem

In summary, conservation of energy problem: Assuming there is no air friction, A 1.4 kg ball is dropped from a height of 2.68 m. It hits the ground, losing 1/12 of its energy in the collision. How high will the ball bounce upward before it comes momentarily to rest? E=K sys + U sysEf=Ei 1/2mv^2+mgh=1/2mv^2+mghI've attempted this problem but I seem to be missing some piece of information and I have run out of ideas hopefully someone can point me in the right direction. Since there are no non-conservative forces then W non conservative
  • #1
greyradio
12
0
[SOLVED] Conservation of Energy problem

Assuming there is no air friction, A 1.4 kg ball is dropped from a height of 2.68 m. It hits the ground, losing 1/12 of its energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?


E = K sys + U sys
Ef = Ei
1/2mv^2 + mgh = 1/2mv^2 + mgh

I've attempted this problem but I seem to be missing some piece of information and I have run out of ideas hopefully someone can point me in the right direction.

Since there are no non-conservative forces then W non conservative = 0 and there are no external forces so that Work is equal to zero.

change in mechanical E = Ef = Ei

Ef = Ei
1/2 m vf^2 + mghf = 1/2 m vi^2 + mghi

I assumed that the final hieght is zero and the initial height is 2.68m. I also assumed that the initial velocity is 0.

1/2 mvf^2 + mg0 = 1/2 m 0 + mghi
1/2 mvf^2 = 1/2 mghi
1/2 (1.4) vf^2 = (1.4)(9.81)(2.68)

i found vf^2 using kinematic equations = 52.5816 m^2/s^2

36.80712 = 36.80712

Since the final energy is 36.80712 and when it collides with the ground the energy disspates by 1/12.
So :
36.80712/12 = 3.06726

So I made the the inital energy of the first bounce.

Ei = Ef
3.06726 = 1/2mv^f + mgh
3.06726 = 0 + mgh
3.06726 = mgh
3.067/mg = h
.22333 = h

This is what I calculated the height to be but this is wrong.

I'm not sure what I'm missing or if I should be looking at it in a different way.
 
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  • #2
PART 1
Potential Energy
E=mgh
E=(1.4)(9.8)(2.68)
E=36.77N
Energy Lost= 36.77/12=3.06N
Total Energy Left= 36.77-3.06=33.71N
The amount of 33.71N is being carried on to the next step.

PART 2
E=mgh
33.71=(1.4)(9.8)(h)
33.71=13.72h
h=2.45m

I am not sure that my work is correct.
 
  • #3
If it loses 1/12 of it's energy KE in the collision, then it's energy after the collision is (11/12)*KE, isn't it? And while I think you are doing this essentially correctly, you could do it a lot more simply. (2.68m)*m*g=(1/2)*m*v^2=KE. After the bounce the KE is (11/12)*KE. So (x)*m*g=(11/12)*KE, where x is the rebound height. So x*mg=(11/12)*(2.68m)*mg. You don't need to know m or g.
 
  • #4
Raza said:
PART 1
Potential Energy
E=mgh
E=(1.4)(9.8)(2.68)
E=36.77N
Energy Lost= 36.77/12=3.06N
Total Energy Left= 36.77-3.06=33.71N
The amount of 33.71N is being carried on to the next step.

PART 2
E=mgh
33.71=(1.4)(9.8)(h)
33.71=13.72h
h=2.45m

I am not sure that my work is correct.

Dead right. (11/12)*2.68~2.45. Now simplify the calculation.
 
  • #5
Thank you. I did not subtract the dissipated energy from the original. Thanks to all that helped.
 

Related to Conservation of Energy problem

What is conservation of energy?

Conservation of energy is a fundamental law in physics that states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. This means that the total amount of energy in a closed system remains constant over time.

Why is conservation of energy important?

Conservation of energy is important because it helps us understand and predict the behavior of physical systems. It also plays a crucial role in many engineering and technological applications, as well as in environmental and sustainability efforts.

How is conservation of energy applied in real-life situations?

Conservation of energy is applied in many real-life situations, such as in the design of energy-efficient buildings and vehicles, the development of renewable energy sources, and the understanding of natural phenomena like weather patterns and the water cycle.

What are some examples of conservation of energy in action?

Some examples of conservation of energy in action include a pendulum swinging back and forth, a roller coaster moving along its track, and a ball bouncing on the ground. In all of these cases, the total energy of the system remains constant, even as the energy is transferred between different forms.

What are the potential consequences of not following the principle of conservation of energy?

If the principle of conservation of energy is not followed, it can lead to unexpected and potentially dangerous outcomes. For example, a machine that does not conserve energy may overheat or malfunction, and a system that does not conserve energy may experience sudden and unpredictable changes in its behavior.

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