Work done by the force of a man on a crate of 80N

[Kennedy]

Homework Statement

A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30° with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2 , then the work done by the man is: (a) –200 J (b) 61 J (c) 140 J (d) 200 J (e) 260 J

Homework Equations

W = Fdcosθ
F = ma

The Attempt at a Solution

I started by calculating the force down the slope of the crate due to gravity, which will be sin30(80) = 40 N. Then I determined the net force acting on the crate. Because the acceleration of the crate is -1.5 m/s^2, F = 8.16(-1.5) = -12.24 N. I calculated the mass by taking 80/9.8 = 8.16 kg. Now, I've done this question several times, but I always mess up the signs, but I still don't understand why some quantities are positive and negative. The speed of the crate is decreasing at a rate of -1.5 m/s^2, so the net force should be in the direction opposite to its direction of motion, right? That means that the net force would be acting up the incline. This yields the force of the man to be 12.24 + 40 = 52.24 N. Then, the work done by the man is 52.24(5) = 260 J, but this is wrong. why am I having so many problems with my signs?

 

[Chandra Prayaga]

All confusions of sign arise because of lack of a proper diagram, with coordinate axes shown. Please draw a clear free body diagram and then we can easily sort out the signs.

 

[Kennedy]

All confusions of sign arise because of lack of a proper diagram, with coordinate axes shown. Please draw a clear free body diagram and then we can easily sort out the signs.

I don't know how to add a picture into my reply, but I do have a sign diagram in front of me. I had down the incline to be positive, and up the incline to be negative.

 

[gneill]

I don't know how to add a picture into my reply,...

If you can create a picture on your computer (say jpg or png or pdf), then you can use the UPLOAD button to upload the file and attach it to your post.

 

[Chandra Prayaga]

The speed of the crate is decreasing at a rate of -1.5 m/s^2, so the net force should be in the direction opposite to its direction of motion, right? That means that the net force would be acting up the incline.

If i put the two parts of your statement together, it looks like you feel that the direction of motion is downward. Why is that so?

 

[Kennedy]

If i put the two parts of your statement together, it looks like you feel that the direction of motion is downward. Why is that so?

Well, contrary to that, what indication do I have that the motion is up the incline? ...and why would the direction of the motion be affected by my sign diagram. Doesn't it not matter which way is positive, and which way is negative? How do I know that the man's force isn't greater than the crate. What if it was decelerating up the incline? What indication do I have that it's not the case?

 

[haruspex]

what indication do I have that the motion is up the incline?

This:

5.0 m upward

Doesn't it not matter which way is positive, and which way is negative?

You can choose either as long as you are consistent. He pushes up and the displacement is up, so they must have the same sign.

How do I know that the man's force isn't greater than the crate.

Because the movement is up but the speed is slowing.

 

[Kennedy]

This:You can choose either as long as you are consistent. He pushes up and the displacement is up, so they must have the same sign.

Because the movement is up but the speed is slowing.

So if the crate were moving up the incline, the crate would have a positive acceleration? That makes sense. But since it's slowing down, the man is able to apply enough force to slow the crate down, but not enough to completely change the direction of the crate. Is this correct?

 

[haruspex]

So if the crate were moving up the incline, the crate would have a positive acceleration?

No, that's not what I wrote. The displacement is up, so the velocity is (on average) up. Since the speed is reducing always it never changes direction, so the velocity is up always. If speed is reducing then velocity and acceleration have opposite sign, so the acceleration is down.

 

[Kennedy]

No, that's not what I wrote. The displacement is up, so the velocity is (on average) up. Since the speed is reducing always it never changes direction, so the velocity is up always. If speed is reducing then velocity and acceleration have opposite sign, so the acceleration is down.

Okay, I think I understand. In this problem the crate is definitely slowing down. Which means that it's direction of travel is opposite of the direction of the acceleration. If the acceleration in the problem is given as being negative, then the direction of travel has to be positive. So, if I have down the incline to be negative, and up to be positive, the crate is moving upwards, while the acceleration is acting downwards, which also means that the net force is acting down, which also implies that the force of the man is less than the force of the crate that is directed down the incline. Now do you think I have the idea?

Conversely, if I have up the incline to be negative, and down to be positive, the direction of travel is down the incline, and the net force acts up the incline, but then this would mean that the force of the man is greater than the force of the crate that acts down the plane. This second part is where I'm getting confused. What if I chose a slightly different sign diagram?

 

[haruspex]

if I have down the incline to be negative, and up to be positive, the crate is moving upwards, while the acceleration is acting downwards, which also means that the net force is acting down, which also implies that the force of the man is less than the force of the crate that is directed down the incline

Yes.

if I have up the incline to be negative, and down to be positive, the direction of travel is down the incline,

Not sure what you are considering here. If you mean the same question but with the other sign convention then the direction of travel is still up the incline. The reversed sign convention will make it a negative value.
Or are you considering a different problem scenario here?

 

[Kennedy]

Yes.

Not sure what you are considering here. If you mean the same question but with the other sign convention then the direction of travel is still up the incline. The reversed sign convention will make it a negative value.
Or are you considering a different problem scenario here?

I'm considering the same problem, but with a different sign convention. Why would the direction of travel still be up the incline? If the acceleration is negative, then the net force is also in the negative direction, opposite to the velocity which would still have to be positive.

 

[Kennedy]

I'm considering the same problem, but with a different sign convention. Why would the direction of travel still be up the incline? If the acceleration is negative, then the net force is also in the negative direction, opposite to the velocity which would still have to be positive.

Ooohhh... I'm missing I very crucial part of the problem here. It's a failure to read properly. The man pushes UPWARDS. Without that indication in the problem, it could have gone either way, correct? I'm not being completely crazy, am I?

 

[haruspex]

Without that indication in the problem, it could have gone either way, correct

Right. (Did you miss the first part of my post #7?)

 

[Kennedy]

Right. (Did you miss the first part of my post #7?)

Yes, I definitely did, and I missed it in the question every single one of the of times that I did the question! Thanks for your help.