Calculating Work Done on a Ball: Where Did I Go Wrong?

[rum2563]

Homework Statement

A 125g ball is thrown with a force of 85.0 N that acts through a distance of 78.0cm. The ball's velocity just before it is caught is 9.84m/s.

Calculate the work done on the ball.

Homework Equations

W= F X d (delta displacement)

The Attempt at a Solution

W = 85N X 0.78m
= 66.3 J

But the answer in my physics textbook is 6.24 J. Can anyone please tell me what I am doing wrong?

 

[Doc Al]

You found the work done by the applied "throwing" force. But perhaps they want the total work done on the ball due to all forces acting on the ball (like air resistance) up to just before it's caught. How would you calculate that?

 

[rum2563]

I would say that I would also have to find the force of gravity, because that is also acting on the object.

Fg = mgh (delta height)
= 0.125kg X 9.81 X 0.78m
= 0.956475 J

But even if I add this to the value of 66.3 J, I still wouldn't get the right answer.

 

[Doc Al]

That's the hard way. (And what about air resistance?) Hint: Take full advantage of the information given.

 

[rum2563]

I am really not sure what to do. I am guessing that maybe I have to find the kinetic force? I am not sure, can you please elaborate more. Thanks.

 

[D H]

What is the relationship between work and energy?

 

[rum2563]

Well, since the chapter is about Power, it says in my book that power is work divided by change in time, and power could also be energy divided by change in time.

Also, I know about how when work is done, there is a change in energy.

 

[Doc Al]

Also, I know about how when work is done, there is a change in energy.

Now you're getting warm!

Express that mathematically.

 

[rum2563]

W = E (delta energy)

i remember reading in my book that work is equal to change in kinetic energy or gravitational potential energy. But I am not sure whether that's right.

 

[Doc Al]

Look up the "Work-KE" theorem (or Work-Energy Principle) in your text. It states that the net work done by all the forces acting on a body equals the change in kinetic energy.

Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr"

 

[Mathgician]

W = E (delta energy)

i remember reading in my book that work is equal to change in kinetic energy or gravitational potential energy. But I am not sure whether that's right.

You are correct, but who is doing the work the system or is energy being transfered(work done by outside force) to the system, or is the system losing potential energy, and gaining negative work? So it is a matter of perspective, you can have positive work and negative work. If the system gains potential energy, there was positive work done, but the system is losing energy gained by work(negative work is being done) if it is transfer it via kinetic or other forms,friction etc...

make sure you pay attention to what object or system they are referring to when talking about work, it is very important! Get familiarized with what system means and how energy is transfered. Work has big part in it, its a term used most often.

 

[rum2563]

ya, but you see the work-energy theorem states that there must be a second kinetic energy and potential energy. the problem with this question is that the final velocity is 0, so that would mean the whole final energy would be 0.

here are the calculations that i am trying to do:

KEi + PEi + Wext = KEf + PEf

(1/2)(0.125)(9.78^2) + (0.125)(9.81)(0.78) = (1/2)(0.125)(0^2) + (0.125)(9.81)(0.78)

5.978025 + 0.956475 = 0 + 0.956475

Now I am not sure what to do, because this formula doesn't lead me to anything.

 

[Doc Al]

ya, but you see the work-energy theorem states that there must be a second kinetic energy and potential energy. the problem with this question is that the final velocity is 0, so that would mean the whole final energy would be 0.

Find the total work done on the ball, by all forces, up until the moment just before it's caught. What's the change in KE of the ball?

 

[rum2563]

To find change in KE, I did:

PEi

(0.125)(9.81)(0.78)
= 0.956475

KEi

(1/2)(0.125)(9.78^2)
= 5.978025

Total Change in KE : 5.978025 - 0.956475

= 5.02155 is the total change in KE.

I hope I am on the right track. Please tell me. thanks.

 

[rum2563]

ya, i am sorry i can't understand because the thing is that the formula for KE is:

(1/2)mv^2

Now, how can there be a change in kinetic energy while nothing is changing? because i am confused since velocity is not changing, so how can there be a change in KE.

 

[Doc Al]

To find change in KE, I did:

PEi

(0.125)(9.81)(0.78)
= 0.956475

On what basis are you calculating PE? You are given no information about the height or change in height. (0.78 m is the distance over which force acts, not the change in height.)

Forget PE. Find the change in KE.

KEi

(1/2)(0.125)(9.78^2)
= 5.978025

Use the correct "final" speed.

 

[Doc Al]

ya, i am sorry i can't understand because the thing is that the formula for KE is:

(1/2)mv^2

Now, how can there be a change in kinetic energy while nothing is changing? because i am confused since velocity is not changing, so how can there be a change in KE.

What makes you think velocity is not changing? Someone just threw a ball! (The initial speed is zero.)

 

[rum2563]

oh my bad,

i was using the wrong speed.
ok, i recorrected my calculation:

KEi
(1/2)(0.125)(9.84^2)
= 6.0516

ok, so the initial speed is zero, that would just mean that the kinetic energy in the beginning is also zero, since the formula for kinetic energy is all multiplication and when we multiply anything by zero, it's zero.
(1/2)(0.125)(0^2) = 0

so, i am thinking that 6.05 is the change in kinetic energy.
is that right? please help, thanks very much.

 

[Doc Al]

so, i am thinking that 6.05 is the change in kinetic energy.
is that right?

That's what I would say. (6.05 Joules.)

 

[rum2563]

ya, but isn't that the kinetic energy? and the question is asking about the work being done on the ball. so that means the kinetic energy and the work being done are actually the same values?

i guess maybe the textbook has the wrong answer. but it couldn't be wrong since it asks about how much energy was lost in the other part of the question.

 

[rum2563]

i am starting to worry because that is not the right answer. i mean work should be 6.24 J and it is not.

 

[denverdoc]

I'm equally confused, the work done on the ball should be the difference between the initial and final kinetic energy. so this way out of ballpark, since the initial Ke was 66.3J and the final 6.05, I'd be thinking the work is the difference which isn't even close to what you posted as an answer. In the meantime, i'd try not to worry, (Something is amiss here over which you have no control).

 

[Doc Al]

ya, but isn't that the kinetic energy? and the question is asking about the work being done on the ball. so that means the kinetic energy and the work being done are actually the same values?

Sure that's the kinetic energy of the ball. The problem is that the question "What is the work done on the ball?" is incredibly vague. Work done by what forces? All the forces? The force of the hand accelerating the ball? Air resistance? The force of the hand catching the ball?

Given such a vague question, you have to guess at what they mean. I guessed that they want the total work done on the ball from the moment the thrower touches it to the moment just before it's caught. (Of course, I'm cheating a bit, since I know that that gives an answer close to what the book gave.)

i guess maybe the textbook has the wrong answer. but it couldn't be wrong since it asks about how much energy was lost in the other part of the question.

Can you post the entire problem exactly as given? Again, energy lost due to what? Since you're given the throwing force, you can calculate the ball's energy immediately after it's thrown. And, since you are given the speed just before impact with the catcher, you can calculate the ball's energy at that moment.

Given no hint at an answer, I would naturally interpret the first question as asking about the amount of work done by the hand. That's easy to calculate. (But does not match the book's answer--too bad!)

And I would naturally interpret the "energy lost" as being the energy lost between the time the ball left the hand and the time just before the ball was caught. Again, easily calculated.

The best you can do is thoroughly understand everything one can calculate about this problem and not worry too much about the book's answer.

Post the complete problem; perhaps the context of all the questions will make things clearer.

 

[rum2563]

thanks for all the help guys. Our physics teacher finally took up the problem in class and it was actually the basic formula for work (force X displacement). She told us there was an error in the book.
I am so mad because i spend so much time thinking maybe there was something missing. Anyways, still thanks for the help guys, even though you guys tried harder to help me than the problem was worth. thanks.

 

[denverdoc]

I know politics are off limits here, but I would ask a favor in return. Write both your school board and the publisher about your experience.

It might be a fine text with a single error, (which would actually be great) or some low cost text w/o math proofing. This would return the favor in abundance and do a great service for all students.