# And yet another ball problem.

Discussion in 'Physics' started by Voltboy, Aug 7, 2008.

1. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
I was at the gym with that medicine balls, I saw this one of 4.56kg (weird weight huh) and just wondered, how much Force, Energy, and Power I need to throw it up to 1.23m high.
After thinking a little a came up with doing the following:
9.81m/s$^{2}$+(9.81m/s$^{2}$*1.23m)= 21.8763m/s$^{2}$.
Then 21.8763m/s$^{2}$*4.56kg= 99.755928N

Is that right??

2. ### silvrstring Active Member

Mar 27, 2008
159
0
I don't think that is right. Have you tried solving for energy first, and then solving for initial velocity?

3. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
Yes but to solve for energy you need joules, and to get joules you need newtons, so i thought i need to get the force first.

4. ### triggernum5 Active Member

May 4, 2008
216
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This is a sneaky way to get us to do your homework right?? <JK>

5. ### RiJoRI Well-Known Member

Aug 15, 2007
536
26
4.56kg = 10 lbs.

--Rich

6. ### Ratch New Member

Mar 20, 2007
1,068
4
Voltboy,

Since you are in a constant gravity field, the velocity will be v^2 = 2*g*h. In other words, a weight dropped a height of h will reach the floor at a velocity of v. Likewise, if you throw the weight from the basement and it attains a velocity of v when it reaches the beginning of the first floor, it will soar to a height of h above the floor.

The energy measured from the first floor is E = w*h, where w is the weight of the object. That is called its potential energy and you will have to supply its equivalent in kinetic energy for the weight to reach height h.

The force depends on how fast you accelerate the weight from the basement to the first floor.

The power depends on how fast you accelerate the weight to attain the velocity required.

Ratch

Last edited: Aug 17, 2008
7. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
I wish I did this kind of stuff in school.. Im still doing the boring simple kinematics.
Weird that it said 4.56kg instead of 10lbs..
So back to the ball question, the "basement" is when I got the ball right over my head, and the "first floor" is when my arms are fully stretched up. So force = the time it takes me to stretch my arm, and the power is how fast I make the ball reach the velocity, but both time are equal for me because the ball will reach that velocity when I finish rising my arms.

8. ### Ratch New Member

Mar 20, 2007
1,068
4
Voltboy,

No, force has nothing to do with time. No, power is not the time it takes to reach a velocity. You know the length of your arms. You know how much kinetic energy you must give to the ball. You know the weight of the ball. You should be able to figure out the average rate of power and the force your arms need to exert.

Ratch

Last edited: Aug 17, 2008
9. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
I didn't knew this was that complicated, but I think I got it now.
Lets say that my arm is 0.5m and it takes me 1/3s to stretch it.
0.5m=(1/2)*a*(1/9)s^2
a=9m/s^2
So, (9m/s^2)*4.56kg= 41.04N, and 41.04N*0.5m=20.52J. If I understood you, I need to exert the potential energy in kinetic energy, so:
E=w*h
E=(9.81m/s^2)*4.56kg*0.5m
E=22.36J
Then 22.36J+20.52J=42.88J, and thats the work I need to do in order to the ball reach 1.23m high.

I hope I did it right.

Thanks for your help Ratch.

10. ### Ratch New Member

Mar 20, 2007
1,068
4
Voltboy,

Looking at the problem again, it appears you don't have to worry about the kinetic energy (KE) since it is the same as the potential energy (PE).

So PE = mgh = 4.56*9.8*1.23 = 54.97 joules

f*0.5 = 54.97 ===> f = 109.3 newtons supplied by you

power = 54.97/(1/3) = 164.9 watts is average power rate

Ratch

Last edited: Aug 17, 2008
11. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
That sounds logic..
Thanks a lot for your help.

While doing this problem I came up with some questions.
If an object going at a velocity of 1m/s (no acceleration) crashes a wall, according to the math then it will exert no force on the wall because f=m*a and as a=0 then f = 0. That sounds a bit illogical for me, and I know that object will exert force on the wall, but how to do it math-wise.

12. ### Ratch New Member

Mar 20, 2007
1,068
4
Voltboy,

You need to read a good physics book about impulse, momentum, elastic, and inelastic collisions. Since the velocity of the object has changed, of course there will be acceleration and force involved.

Ratch

Last edited: Aug 17, 2008
13. ### SIcam Active Member

Aug 9, 2008
61
0
f=m*a

Acceleration (a) is the time rate of change of velocity. The balls velocity will change therefore a force will be excerted.

If the wall is made of 1 piece thick of tissue paper the force exerted will be little and the ball will have velocity after breaking throught the wall. It the wall is brick the force will be large as its velocity will have to stop and change direction.

The internet is a good place to get some good education on physics.