Calculating Work: A Simple Yet Tricky Problem

[fomenkoa]

This seems easy but I'm not getting the right answer!

The question: A 0.5 kg ball is thrown into the air. At a height of 20m above the ground, it is traveling at 15 m/s.

a)How much work was done by someone at ground level throwing the ball up into the air


Ok, for a), Work=Force x Distance
so,

W = (0.5 kg x 9.81 N/kg) x 20 m
= 98 J

But that's not the answer in my textbook! I realize my substitution for "Force" is wrong but I don't know why

Anton

 

[dextercioby]

That work (which is asked) is nothing but the initial KE...Find it.

Daniel.

 

[Lucretius]

What is the answer your book got?

Wouldn't your force be 0.5kg x 15m/s^2=7.5N

7.5N x 20m=150J

*Note, just a guess. I probably don't know the right answer either. I really should have taken a physics class this year.

Just want to try to be helpful.

 

[Jameson]

Another thing... remember that once the person's hands leave the ball, the only force acting on it is gravity (and air resistance, but that's negligable). Remember the nice rule that \bigtriangleup{K.E.} = W

 

[fomenkoa]

What is the answer your book got?

Wouldn't your force be 0.5kg x 15m/s^2=7.5N

7.5N x 20m=150J

*Note, just a guess. I probably don't know the right answer either. I really should have taken a physics class this year.

Just want to try to be helpful.

Yes! The answer is 150 J! However, Force=mass x accel
You said that the accel is 15 m/s/s...why?? Isnt that the speed? Why is the speed and accel the same?

 

[dextercioby]

Shouldn't it be 156.25 \ \mbox{J}...?

Daniel.

 

[Lucretius]

Yes! The answer is 150 J! However, Force=mass x accel
You said that the accel is 15 m/s/s...why?? Isnt that the speed? Why is the speed and accel the same?

I got something right? Wow! Confidence boost.

15m/s would be the speed. However, 15m/s^2 would be acceleration.

Speed can be meters per second, but acceleration the change of the the rate of speed, which would be 15m/s^2.

 

[fomenkoa]

Shouldn't it be 156.25 \ \mbox{J}...?

Daniel.

My book says 1.5e2 J...I guess that's to 2 significant digits...but can anyone explain to me why the 15 m/s became a 15 m/s/s?

 

[fomenkoa]

I got something right? Wow! Confidence boost.

15m/s would be the speed. However, 15m/s^2 would be acceleration.

Speed can be meters per second, but acceleration the change of the the rate of speed, which would be 15m/s^2.

But...to know accel, wouldn't you need the initial velocity which we don't know? I still don't understand

 

[dextercioby]

If you mean 1.5 \cdot 10^{\mbox{2}} ,then i believe you.

You have to realize that u're being asked to find the work done to imprime a certain velocity to the ball.And that work is exactly the initial KE of the ball.U'll have to use the theorem of varation of KE (or the law of the total mechanical energy) to find it.

It's not difficult at all.

Daniel.

 

[fomenkoa]

If you mean 1.5 \cdot 10^{\mbox{2}} ,then i believe you.

You have to realize that u're being asked to find the work done to imprime a certain velocity to the ball.And that work is exactly the initial KE of the ball.U'll have to use the theorem of varation of KE (or the law of the total mechanical energy) to find it.

It's not difficult at all.

Daniel.

Oups yeah I meant 1.5e2 ...so

Ek = 1/2 x m x v^2
I know the mass, but not the initial velocity!

 

[Lucretius]

But...to know accel, wouldn't you need the initial velocity which we don't know? I still don't understand

A=\frac{\delta v}{\delta t}.

So your book says 15m/s, and not 15m/s^2?

 

[jdavel]

Are you sure thye answer is supposed to be 150J? I get alittle over 154J.

 

[fomenkoa]

A=\frac{\delta v}{\delta t}.

So your book says 15m/s, and not 15m/s^2?

Right. It says that at 20m, it is traveling at 15 m/s

 

[fomenkoa]

Are you sure thye answer is supposed to be 150J? I get alittle over 154J.

It's to 2 significant digits (rounding)
Oh boy I still don't know how to solve this problem

 

[dextercioby]

Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.

 

[Lucretius]

Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.

what'd I do wrong?

 

[fomenkoa]

what'd I do wrong?

That's what I want to know too,,I still don't get it

 

[dextercioby]

There's no indication of any force acting on the ball during flight,therefore,it's natural to assume that the only force doing work on the ball while in motion is the force of Earth's gravitational attraction.

Daniel.

 

[jdavel]

fomenkoa,

What Lucretius did to solve this makes no sense; he lucked out getting close to the right answer (no offense Lucretius!).

Dextercioby is steering you the right way, pay attention to him.

 

[Lucretius]

I guess I shouldn't attempt to help people anymore, heh.

Alright, bye.