How Is Work Calculated in These Two Physics Problems?

[Underdog_85]

Two Work problems I can't figure out. I don't really know where to start?
Question1) A delivery person carries a 215N box up stairs 4.20m vertically and 6.80m horizontally.
a)How much work does the delivery person do?(Answer: 903 J)
b)How much work does the delivery person do in carrying the box down the stairs?(-903J)

Question2) A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?(Answer: 10 700J)

Can someone explain in full detail how to reach their answers? I tried them and get nowhere close.

Relevant Equations
W=Fs
W=Fscostheda
W=fs=mgs

I know that in Q1) a) if you go W=(215N)(4.20m) = 903J However, what do we do about the 6.80m? Also.. Going down the stairs becomes negative 903J? Is that becomes they go downwards with the box? I really don't know how to properly display my answers for this.

 

[srmeier]

Let's start with problem #1:

Would you not agree that because the box has no velocity at the top of the stairs the total work done on the box is zero? Also, which forces do work on the box?

When can work be negative? Only when the force is opposite the displacement.

 

[Underdog_85]

Let's start with problem #1:

Would you not agree that because the box has no velocity at the top of the stairs the total work done on the box is zero? Also, which forces do work on the box?

When can work be negative? Only when the force is opposite the displacement.

I wouldn't agree with you. If the displacement is 0 than the work would be 0, however there is a displacement of 4.20m, plus the final answer is also 903 Joules or Newton Metres in the back of the text. So the answer cannot be 0.

 

[srmeier]

I'm not saying the answer is zero. We are being asked for the work done by the man; I'm saying the total work done is zero.

\Sigma W=fd=\Delta Ke=0 -(at the top of the stairs)

Now which forces do work on the box?

 

[berkeman]

I wouldn't agree with you. If the displacement is 0 than the work would be 0. The final answer is also 903 Joules or Newton Metres. So the answer cannot be 0.

What force opposes the vertical movement of the box? So how much work is done in that part? What force opposes the horizontal movement of the box? How much work is done in that part?

 

[berkeman]

I'm not saying the answer is zero. We are being asked for the work done by the man; I'm saying the total work done is zero.

W=fd=\Delta Ke=0 -(at the top of the stairs)

Now which forces do work on the box?

Is the delta-KE meant to indicate some kinetic energy component in this question?

 

[srmeier]

Is the delta-KE meant to indicate some kinetic energy component in this question?

I'm saying that:

Ke_f - Ke_i = 0 = \Sigma W

because the box starts at rest and finished at rest the total work done on the box better be zero.

 

[berkeman]

I'm saying that:

Ke_f - Ke_i = 0 = \Sigma W

because the box starts at rest and finished at rest the total work done on the box better be zero.

Not so. The OP could use PE changes to calculate the work done, or multiply the force by the distance through which the force was exerted. KE has nothing to do with this problem, and the work done on the box is definitely not zero.

 

[srmeier]

Not so. The OP could use PE changes to calculate the work done, or multiply the force by the distance through which the force was exerted. KE has nothing to do with this problem, and the work done on the box is definitely not zero.

Total work done = 0

the man does positive work and gravity does negative work of equal magnitude. Thus resulting in zero net work on the box.

 

[berkeman]

Total work done = 0

the man does positive work and gravity does negative work of equal magnitude. Thus resulting in zero net work on the box.

No, srmeier, that is just plain incorrect. I don't want to give too much away here in this thread, because we will end up answering the OP's question for him. We should probably carry this on via PM. So PM me what the delta-PE is for the box in the movement described above, and tell me where that change in potential energy came from. Thanks.

 

[Underdog_85]

Ok I am getting two different answers and it is just confusing me. This is what I did for 1 a)
W=Fs
W=(215N)(4.20m)
W=903Nm or J

1 b)
W=Fs
W=(-215N)(4.20m)
W=-903Nm or J

These are both the final answers in the book, however I am still confused about what we do with the horizontal 6.80m.

 

[Underdog_85]

Question 2
W=Fs(costheda)
W=(200N)(25.0m)(cos25.0degrees)
W=4531 N

This is incorrect though.. I don't know.. This is just frustrating me now. They don't give examples for this in the book.

 

[berkeman]

Ok I am getting two different answers and it is just confusing me. This is what I did for 1 a)
W=Fs
W=(215N)(4.20m)
W=903Nm or J

1 b)
W=Fs
W=(-215N)(4.20m)
W=-903Nm or J

These are both the final answers in the book, however I am still confused about what we do with the horizontal 6.80m.

Good. There is no force opposing the horizontal motion, so there is no work done horizontally in this problem.

 

[berkeman]

Question 2
W=Fs(costheda)
W=(200N)(25.0m)(cos25.0degrees)
W=4531 N

This is incorrect though.. I don't know.. This is just frustrating me now. They don't give examples for this in the book.

Hint -- if he pushes 25m total at an angle of 25 degrees, what is the total vertical displacement?

 

[Underdog_85]

I don't know how to get the vertical.. There is no example.

Would I use trig to find the vertical? hypotaneuse being 25, inside angle being 25, and figure out the opposite side(vertical)?

 

[berkeman]

I don't know how to get the vertical.. There is no example.

Would I use trig to find the vertical? hypotaneuse being 25, inside angle being 25, and figure out the opposite side(vertical)?

Yep.

 

[Underdog_85]

Yep.

I got 10.25m for the vertical..

I popped that in the W=Fs and W=FsCostheda equations but I still don't get 10 700J.

 

[berkeman]

I got 10.25m for the vertical..

I popped that in the W=Fs and W=FsCostheda equations but I still don't get 10 700J.

Yeah, not sure why that isn't giving the answer they provide. Does the problem say anything else? Is it copied into your post correctly?

 

[srmeier]

I must dispute berkeman's method.

"A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?"

He is pushing the wheel barrow in the horizontal direction (not up an incline). the vertical displacement is zero. Although, the man does do work.

(note: even with my changes the answer isn't 10,700J. Also, if the wheel barrow weighted twice as much and had a vertical displacement of 25m the work done by gravity still wouldn't reach 10,700J. Please check your book.)

 

[Underdog_85]

I must dispute berkeman's method.

"A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?"

He is pushing the wheel barrow in the horizontal direction (not up an incline). the vertical displacement is zero. Although, the man does do work.

(note: even with my changes the answer isn't 10,700J. Also, if the wheel barrow weighted twice as much and had a vertical displacement of 25m the work done by gravity still wouldn't reach 10,700J. Please check your book.)

I got 4300J. It does say 10700J in book. It is mistake.

 

[PhanthomJay]

I got 4300J. It does say 10700J in book. It is mistake.

I don't see how anyone can get any answer in such a poorly worded problem.

 

[berkeman]

Without friction, the answer should be mgh. Don't see how it could be different.

 

[PhanthomJay]

Without friction, the answer should be mgh. Don't see how it could be different.

Yes, that would be true if the problem was worded this way:

A laborer pushes a wheel barrow weighing 200N up a ramp inclined at 25.0 degrees above the horizontal. If he pushes it with a force, parallel to the incline, a distance of 25m up the ramp, how much work is done by the laborer, asuming it is pushed at constant velocity? Neglect any friction between the wheelabarrow and the ramp.

The actual problem is worded :

A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?

That problem statement is missing lots of info. For one thing, it says 'how much work is done?' Is this asking for work done by the laborer, or is it asking for net work done by all forces ? (net work done by all forces, (worker force and gravity force) is 0 if the wheelbarrow is pushed at constant speed (also not noted)). And one must assume that it is being pushed up a ramp, it would be nice if that was so stated. A picture would surely have helped.