Kinetic Energy of 7.0 kg Bowling Ball Falling from 2.0 m High Shelf

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A 7.0 kg bowling ball falling from a 2.0 m shelf will convert its potential energy into kinetic energy just before impact. The potential energy (PE) at the height is calculated as PE = mgh, resulting in approximately 137.2 J. This potential energy is equal to the kinetic energy (KE) just before hitting the ground, meaning KE is also 137.2 J. The discussion emphasizes using the correct equations to find velocity and energy, highlighting that all potential energy transforms into kinetic energy during the fall. Understanding this energy transformation is key to solving the problem accurately.
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A 7.0 kg bowling ball falls from a 2.0 m high shelf. Just before hitting the floor, what will be its kinetic energy? (g = 9.8 m/s² and assume air resistance is negligible)

a. 14 J
b. 19.6 J
c. 29.4 J
d. 137 J

Now I tried using the normal Kinetic Energy KE equation KE = 1/2mv² but when I plug 7.0 in for mass, and (9.8 m/s²)(2.0m) for velocity and square it, my answer is much larger than any of the choices.

Like 4,705 so I figure I am doing something wrong.
 
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i am guessing there's no initial velocity.
use the equation that has all the knowns and final velocity to find out what the velocity is before hitting the floor.
with that velocity, you can use KE equation.
my answer is one of those and you should get the same
good luck!
 
There is an easier way to do that.
What happens to all the potential energy that your system has initially when falling?
 
krnhseya

So am I using the wrong equation? What equation do you mean that has all of the knowns?

<---
I am up for an easier way if there is one. Doesn't all of the potential energy change to kinetic engergy since it is falling and actually moving?
 
21ducks said:
krnhseya

Doesn't all of the potential energy change to kinetic engergy since it is falling and actually moving?

That's correct. So do you know the equation for gravitational potential energy on earth?
You had the right idea at first I think, but velocity does not equal ad.
 
yeah you are on right track and the equation that i was talking about is...

V^2 = Vo^2 - 2g(change in height)

that will get you V and you can use KE.
 
I think I found it in my book. Is it PE = mgy ?

y = 0 right since it is the ground? But won't that just make potential energy equal to 0? I guess I am still a little confused. Thanks for all the help though.
 
PE = mgh

you are asked to find KE before it hits the ground so you are looking for PE at initial so y (or h) is NOT 0 ;)

increase in PE, decrease in KE
increase in KE, decrease in PE
 
Okay so PE = 137.2 right. Now that I know the PE how do I use the equation you mentioned above to obtain the KE?

The one I am referring to is: V^2 = Vo^2 - 2g(change in height)?
 
  • #10
21ducks said:
Okay so PE = 137.2 right. Now that I know the PE how do I use the equation you mentioned above to obtain the KE?

The one I am referring to is: V^2 = Vo^2 - 2g(change in height)?

yeah you get the samething at the end.
 
  • #11
So are you saying that the PE = KE? Which would mean KE = 137.2?

Just out of curiosity if that is so, what does Vo^2 = ?
 
  • #12
Yes, in the end your initial potential energy will equal your final kinetic energy.

And if you want to find v_f now, you need only to solve for it in your kinetic energy equation, but you can still just use the equation krnhseya posted.

v_f^2 = v_i^2 + 2a\Delta{y}

Alternatively you could also get velocity by solving using,
s = v_it + \frac{1}{2}at^2
v_f = v_i + at
 
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