Question

A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of 35.9 above the horizontal. If the monitor's speed is a constant 2.40 , how much work is done on the monitor by friction, gravity, and the normal force of the conveyor belt?

Answer 1

For constant forces, like you have, the work is defined as follows.\[W=\vec F \cdot \Delta \vec r \equiv F \Delta r\cos \theta\], where \(\cdot\) denotes the dot product. \(\Delta \vec r\) is the displacement, and the alternate form uses \(\theta\) which is the angle between the force and the displacement. Do this three times, once for each force. (Hint: one of the work values will be zero since you will find \(\theta=90^\circ \)). Let me know where that takes you.

Answer 2

Hint: since \(\Delta \vec r\) is the displacement of the object, and you're just interpreting the work three times, \(\Delta \vec r\) won't change in your three calculations.

Answer 3

ok let me work it Ill be right back :)

Answer 4

so for work of gravity I did this:
F[\Delta r\]cos= (12)(9.8)(7.5)cos35.9, is this right?

Answer 5

Very close... the angle you need is actually the compliment of that. Gravity points down, while the object goes up on a diagonal, so the angle needs to be somewhere between \(90^\circ\) ad \(180^\circ\).

Answer 6

compliment you meant closest to Y,...54.1

Answer 7

Whoops sorry....... SUPPLEMENT is what I meant....though I actually realized that that's not quite either. The angle is in fact still between 90 and 180 though...look at the picture I drew...what is the angle between the vectors? ;)

Answer 8

The up-right arrow is \(\Delta \vec r\) and the down arrow is \(m \vec g\).

Answer 9

wooo hooo thank you that remark helped me to solve for gravity, the angle then it is actually 54.1, since gravity is consider negative my answer would be -517.N right?

Answer 10

Im having a hard time with friction :(

Answer 11

The angle is \(\theta = 35.9^\circ + 90^\circ = 125.9^\circ\) by the picture. It will cause cosine to be negative of what you got, which in effect did the same as your instinct to negate the work. Yes, if the object is moving in the opposite direction as the force that is acting on it, then the work is negative.

Answer 12

First off, what direction is the friction force pointing?

Answer 13

that is the hard part for me to see it, since it says that the friction is actually doing the moving :(

Answer 14

Correct. Since friction is ALWAYS along the plane of the surface, then do you figure it's pointing upward or downward?

Answer 15

should be pointing downward right? since is doing the moving...

Answer 16

Which of these free body diagrams do you think more accurately describes a ball that is moving up the ramp? (I only switched the direction of the friction force here)