Physics Problem Work/Energy

[betterfuture]

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A force of 250 N is used to push a 30kg wagon uphill a distance of 12m, bringing it to a point 6 meters higher than its starting point. If 10% of work done is used in overcoming friction,

a)Find the amount of Work Done

I need some help here. Thank you.

 

[aldol16]

Without friction, 250 N is exerted over a distance of 12 m, giving you a work of 3000 J done. That's how much work you do, period. So the answer to (a) should be 3000 J.

But part of that comes from friction. In fact, 10% of that is from friction. So you would only exert 2700 J of work if there had been no friction. But since the question doesn't ask about that, don't worry about it.

 

[betterfuture]

Without friction, 250 N is exerted over a distance of 12 m, giving you a work of 3000 J done. That's how much work you do, period. So the answer to (a) should be 3000 J.

But part of that comes from friction. In fact, 10% of that is from friction. So you would only exert 2700 J of work if there had been no friction. But since the question doesn't ask about that, don't worry about it.

So are you saying the answer to a) is ? I got kind of lost.

 

[theonlytycrane]

does the answer ask for the amount of work done by the force of 250N or total?

 

[betterfuture]

It asks for the amount of work done

 

[aldol16]

So are you saying the answer to a) is ? I got kind of lost.

From the way you word the question, 3000 J.

 

[aldol16]

does the answer ask for the amount of work done by the force of 250N or total?

The work done by 250 N of work and the total work are one and the same by the wording in the question.

 

[betterfuture]

From the way you word the question, 3000 J.

Is it wrong for me to think that after they stated 10% was lost to friction, I got 2700J and not 3000J. They didn't exactly specify which amount of work done they wanted? Couldnt work also be done by friction and answer could also be 300J? There are many different types of work done. Work from non conservative forces, work from change in mechanical energy, work from change in KE. Am I wrong to think they wanted 2700J as the answer?

 

[aldol16]

Is it wrong for me to think that after they stated 10% was lost to friction, I got 2700J and not 3000J. They didn't exactly specify which amount of work done they wanted? Couldnt work also be done by friction and answer could also be 300J? There are many different types of work done. Work from non conservative forces, work from change in mechanical energy, work from change in KE. Am I wrong to think they wanted 2700J as the answer?

From what you posted, I don't think the question is clear with regard to this. However, there's only two types of work they could be mentioning. Work from friction or work from you. Change in mechanical energy isn't work - you're inputting the work to change the potential and kinetic energies.

From my interpretation, they want the total work done by you. Which would simply be 3000 J. But if they want the work done by you that wasn't just to overcome friction, it would be 2700 J.

 

[betterfuture]

From what you posted, I don't think the question is clear with regard to this. However, there's only two types of work they could be mentioning. Work from friction or work from you. Change in mechanical energy isn't work - you're inputting the work to change the potential and kinetic energies.

From my interpretation, they want the total work done by you. Which would simply be 3000 J. But if they want the work done by you that wasn't just to overcome friction, it would be 2700 J.

Ok, so another question was find
d)the increase in KE of the wagon

I worked it out, I just want to see what others get.

 

[aldol16]

You don't have enough information to answer the kinetic energy. It depends only on how fast it's moving when you're pushing it.

 

[betterfuture]

You don't have enough information to answer the kinetic energy. It depends only on how fast it's moving when you're pushing it.

Haha, so would you agree this question is confusing as crap? Book says the answer as 940 J. And here you state there is not enough info given. Thank God I am not alone in this confusion.

 

[aldol16]

Haha, so would you agree this question is confusing as crap? Book says the answer as 940 J. And here you state there is not enough info given. Thank God I am not alone in this confusion.

Yeah, you're not confused. The problem is just not correct as written. Kinetic energy depends only on speed at a certain point in time. So if you're pushing something, the change in kinetic energy depends on when you're assessing it.

 

[betterfuture]

So in case you were wondering, they stated the work produced 3 kinds of energy. W=internal E+KE+PE and since work done was 3000J, when you solve for KE they got 940J

3000J=300J+KE+1760J

 

[aldol16]

I calculated potential energy and I still don't see how they could get that. Potential energy change (relative to before beginning to push) is equal to m*g*h = (30 kg)(10m/s^2)(6 m). That's 1800 J.

 

[betterfuture]

I calculated potential energy and I still don't see how they could get that. Potential energy change (relative to before beginning to push) is equal to m*g*h = (30 kg)(10m/s^2)(6 m). That's 1800 J.

Ahhh, I just caught that. I calculated 1800 J as well but in the back of the book it states 1760J NOT 1800J. I just don't know anymore. This is the one area that you don't want to find mistakes because I am already already having a great deal of tension with physics thus far and now I have to deal with confusing perplexed horrendous physics problems that make no sense.

 

[FutureD0C17]

Physics is a mfer! Just gotta keep working at it. The 1760 comes from gravity being 9.8.

 

[betterfuture]

Physics is a mfer! Just gotta keep working at it. The 1760 comes from gravity being 9.8.

Thank you for catching that. My brain is just out it today with the physics. So it IS 1760J. I am not sure if you followed what the question was, but did you understand how they got the KE? @FutureD0C17

 

[FutureD0C17]

W = change in KE + change PE

We know the net work after friction is 2700J. We also know the change in PE (30kg*9.8*6). The only possible way to calculate KE from this problem is to simply subtract the PE from 2700. Poorly worded question.

 

[aldol16]

We know the net work after friction is 2700J. We also know the change in PE (30kg*9.8*6). The only possible way to calculate KE from this problem is to simply subtract the PE from 2700.

Only possible way? Sure. But legitimate? No. Change in KE? Well, a change implies a difference between two different points. Which points could they be referring to? Well, the only two logical points would be at h = 0 and h = 6. Well, presumably, your 250 N is exerted only over 12 meters and the block isn't moving further (otherwise it would be more overall work). If the block stops at 12 m displacement, then the KE must necessarily be 0 since it's not moving. Just a terrible question in general.

 

[FutureD0C17]

Agreed. Very poor question. Where did this one come from anyway?

 

[betterfuture]

And also terrible for the fact that they stated in the question stem, "if 10% of the work done is used to overcome friction", so here I am confused because then they say to find the total work done. Okay. What work? Work AFTER 10% of it was converted to friction, or OVERALL Work like the work I did to push it AND the 10% that went to friction which would be two totally different answers, or, or, or the work done by non conservative forces which also would yield a totally different answer.

 

[betterfuture]

Agreed. Very poor question. Where did this one come from anyway?

Barron's E-Z Physics. It has a bunch of physic questions, like a mini test at the end of the chapters. I thought it was pretty good but here I got pretty confused with their wording on the question.