Kinetic Energy and bowling ball Question

[JCB]

Homework Statement

A 7kg bowling ball falls from a 2m shelf. Just before hitting the floor, what will its kinetic energy be? (air resistance is negligible)

Homework Equations

PEi + KEi = PEf + KEf
KE = 1/2mv^2

The Attempt at a Solution

m = 7kg
Δy = 2m
g = 9.8m/s^2
Vi = 0 m/s
I solved for the final velocity, using PEi + KEi = PEf + KEf, and found it to be 39.2 m/s. Then, I plugged that into the equation KE = 1/2mv^2, and got the Kinetic Energy, but my answer was way off. (The correct answer is 137J)

 

[PhanthomJay]

Homework Statement

A 7kg bowling ball falls from a 2m shelf. Just before hitting the floor, what will its kinetic energy be? (air resistance is negligible)

Homework Equations

PEi + KEi = PEf + KEf
KE = 1/2mv^2

The Attempt at a Solution

m = 7kg
Δy = 2m
g = 9.8m/s^2
Vi = 0 m/s
I solved for the final velocity, using PEi + KEi = PEf + KEf, and found it to be 39.2 m/s. Then, I plugged that into the equation KE = 1/2mv^2, and got the Kinetic Energy, but my answer was way off. (The correct answer is 137J)

You have the correct relevant equations. But you left out the formula for PE. Define what PE is and try again, please.

 

[JCB]

You have the correct relevant equations. But you left out the formula for PE. Define what PE is and try again, please.

PE = mgy

but I don't need any help on this. I realized what I did wrong: nothing!

I got v^2 to be 39.2, and forgot to solve for just v.

That would explain a lot.

 

[PhanthomJay]

PE = mgy

but I don't need any help on this. I realized what I did wrong: nothing!

I got v^2 to be 39.2, and forgot to solve for just v.

That would explain a lot.

BUT why are you solving for V? you're working in reverse. You have to find the KE first, then you solved for V, then you plugged it back and solved for KE again?

 

[JCB]

BUT why are you solving for V? you're working in reverse. You have to find the KE first, then you solved for V, then you plugged it back and solved for KE again?

Your looking for the KE just before the ball hits the floor. In order to do that, you need to first find it's velocity just before it hits the floor. So you do PEi + KEi = PEf + KEf. KEi goes to 0 since it's starting from rest, and PEf goes to 0 since y=0 just before it hits the floor. So you're left with PEi = KEf, which can also be written as mgy = 1/2mv^2. Plug everything in: (7)(9.8)(2)=1/2(7)v^2

you get v^2=39.2

and me being the lazy guy I am forgot to write v^2 and just wrote v, which meant my final velocity was way off. Once I realized I forgot to square root each side, I was able to get the real final velocity (6.26 m/s) and then plug that into KEf = 1/2mv^2, which means KEf = 1/2(7)(39.2), which means that KEf = 137.2 J.

So, in short, I couldn't solve for KEf to begin with because I didn't have the final velocity to begin with. If I already had the final velocity, that would have been a joke question.