Calculating Work: A Simple Yet Tricky Problem

In summary, the question is asking for the work done by someone at ground level throwing the ball up into the air, and the answer is 150 J. However, you may not have gotten the answer correctly if you substituted the word "force" for "work". Additionally, you should take a physics class to understand the more complicated concepts discussed in this conversation.
  • #1
fomenkoa
47
0
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
 
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  • #2
That work (which is asked) is nothing but the initial KE...Find it.

Daniel.
 
  • #3
What is the answer your book got?

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

[tex]7.5N[/tex] x [tex]20m=150J[/tex]

*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.
 
  • #4
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 [tex]\bigtriangleup{K.E.} = W[/tex]
 
  • #5
Lucretius said:
What is the answer your book got?

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

[tex]7.5N[/tex] x [tex]20m=150J[/tex]

*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?
 
  • #6
Shouldn't it be [itex] 156.25 \ \mbox{J} [/itex]...?

Daniel.
 
  • #7
fomenkoa said:
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.

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

Speed can be meters per second, but acceleration the change of the the rate of speed, which would be [tex]15m/s^2[/tex].
 
  • #8
dextercioby said:
Shouldn't it be [itex] 156.25 \ \mbox{J} [/itex]...?

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?
 
Last edited:
  • #9
Lucretius said:
I got something right? Wow! Confidence boost.

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

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

But...to know accel, wouldn't you need the initial velocity which we don't know? I still don't understand :redface:
 
  • #10
If you mean [tex] 1.5 \cdot 10^{\mbox{2}} [/tex] ,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.
 
  • #11
dextercioby said:
If you mean [tex] 1.5 \cdot 10^{\mbox{2}} [/tex] ,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!
 
  • #12
fomenkoa said:
But...to know accel, wouldn't you need the initial velocity which we don't know? I still don't understand :redface:

[tex]A=\frac{\delta v}{\delta t}[/tex].

So your book says [tex]15m/s[/tex], and not [tex]15m/s^2?[/tex]
 
  • #13
Are you sure thye answer is supposed to be 150J? I get alittle over 154J.
 
  • #14
Lucretius said:
[tex]A=\frac{\delta v}{\delta t}[/tex].

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

Right. It says that at 20m, it is traveling at 15 m/s
 
  • #15
jdavel said:
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 :confused:
 
  • #16
Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.
 
  • #17
dextercioby said:
Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.

:frown: what'd I do wrong?
 
  • #18
Lucretius said:
:frown: what'd I do wrong?
That's what I want to know too,,I still don't get it
 
  • #19
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.
 
  • #20
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.
 
  • #21
I guess I shouldn't attempt to help people anymore, heh.

Alright, bye.
 

Related to Calculating Work: A Simple Yet Tricky Problem

1. What is the definition of work in physics?

In physics, work is defined as the product of force and displacement, where force is applied to an object and causes it to move in the direction of the force.

2. Can work be negative?

Yes, work can be negative if the force and displacement are in opposite directions. This means that the force is acting against the direction of motion, resulting in negative work.

3. Is there a maximum limit to the amount of work that can be done?

No, there is no maximum limit to the amount of work that can be done. As long as there is a force applied and a resulting displacement, work can continue to be done.

4. What is the unit of measurement for work?

The unit of measurement for work is joules (J) in the International System of Units (SI). It can also be measured in other units such as foot-pounds (ft-lb) or ergs (erg).

5. How is work related to energy?

Work and energy are directly related, as work is the transfer of energy to or from an object. The unit of measurement for work (joules) is also the unit for energy, showing their connection in the physics equations.

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