# Help with concepts

1. Feb 1, 2016

### ApPhysicsStudent

• Thread moved from the technical forums, so no HH Template is shown.
14). Student A lifts a 50-N box from the floor straight up to a height of 40 cm in 2 s. Student B lifts a 40 N boxes straight up from the floor to a height of 50 cm in 1 s. Compared to Student A, Student B does

(A) The same work but develops more power.

(B) The same work but develops less power.

(C) More work but develops less power.

(D) Less work but develops more power.

15). Now Student A instead lifts the 50 N box from the floor diagonally, moving the box 40 cm to the right and 40 cm upward in the same 2 s.

(A) Compared to the work he did originally, does Student A do more, less or the same work?

(B) Compared to the power he developed originally, does Student A develop more, less or the same power?

Ok, so I know that on #14 the answer is A, same work through the formula F*D=W.
However, our physics teacher claims that on #15 that the answer is Same work and power. She says its based on the equation M*G*H=W because there is no change in KE. I understand what she is talking about.
But, the way I see it is, you should use the equation F*D*CosX=W. X is theta, which you find by the angle between the direction of the force and the direction of the object; in this case it is 0. So then F*D=W. With Pythagorean Theorem, you find that D=56.56 and Force is given at 50N, for a total of 28.28J and 14.14W.
Which one is correct and why is the other not?

2. Feb 1, 2016

### Aniruddha@94

The purpose of the dot product is :
1) you want the component of the force in the direction of the displacement
2)Another way to think about it is that the dot product gives the amount of displacement in the direction of the force.
In your question 15) since gravity acts downward and there's no force in the horizontal direction ( it means that the man can move the box in the horizontal direction freely)... The only work being done is against gravity

3. Feb 1, 2016

### Aniruddha@94

Think of a figure like stairs.. For the horizontal parts there's no work( since displacement is normal to the gravity).. And all the vertical parts of the work are added up. Hope this helps!

4. Feb 1, 2016

### ogg

It appears that the question is a bit misleading. Depending on how you define work, you may get different answers. Lets take the example and instead of "doing it" in the classroom, lets do it in high-Earth orbit. Now, perhaps you see the problem? You have to accelerate the box (F=ma) so that after 2 seconds it is at the specified position, right? OK, my question is: is it at rest or not? It is not, unless you've deccelerated it (which is not part of the problem). So, the amount of work you have done (even considering the gravitational field constant with height (for small differences in height), depends on the box's final velocity. Not given in the question. But the reason that the question doesn't "go there" isn't because the author(s) forgot, its because they were assuming a definition of work which is the "classical thermodynamic" quantity corresponding to the (minimum) limiting energy required to move the box (which is the same (in a simple first order world) regardless of the time taken to move it (ie the thermodynamic work or reversible work)).

5. Feb 1, 2016

### Nidum

40 N box ? 50 N box ?

Last edited: Feb 1, 2016
6. Feb 1, 2016

### WWGD

Think it is Newtons for mass (I guess Monday mass ;)).

7. Feb 1, 2016

8. Feb 1, 2016

### haruspex

For the vertical lift, that difficulty does not arise. The student could accelerate the box up to such a height and speed that it then just makes it up to the desired height in the desired time with no residual speed.
Less obviously, we might also get around it when there is a horizontal displacement. The student could put work into getting it into some diagonal trajectory, then apply only an orthogonal force to arc it round to finish moving vertically and just coming to rest at the desired position. The details don't matter, provided we can show it is feasible.
I don't have have a problem with that. We commonly speak of an object of weight so-many Newtons. Sure, it depends on the gravitational field, so is not an inherent property of the object.

9. Feb 1, 2016

### ApPhysicsStudent

Thank you guys i understand now but the stairs comment was what clicked for me, and sorry if this was like an easy question i just dont really learn in my classroom because of how it is set up so i dont fully understand every thing yet.

10. Feb 2, 2016

### Aniruddha@94

Don't worry about it. Believe me, you don't learn as much in the classroom as you do by thinking on your own.. School lessons are just a guide to your understanding