# TBR Physics - Ch.3, Passage VI

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#### MelloTangelo

##### Full Member
10+ Year Member
42. How much work is done by friction in the region from point b to point c?
a. mgh
b. mgd(c-b)
c. mgcos(theta)d(c-b)
d. 1/2mv^2

Sorry if that's a little hard to read. The original passage shows a frictionless ramp from a to b, and a flat space of rollers from b to c. The answer choice says something about it being a because the work is the same as lifting the box, but really, their explanation isn't making any sense to me whatsoever.

Anyone gotta' clue?

Thanks!

##### Full Member
10+ Year Member
@MelloTangelo

What do you mean by a "flat space of rollers"?

Like it being horizontal?

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#### Cawolf

##### PGY-2
10+ Year Member
It would be really helpful if you could post a page number next time.

The passage states that the box comes to rest at point c, and the friction is applied by the rollers starting at point b.

So gravity does mgh work on the box as it slides to base of the inclined plane, and then the rollers bring it to a stop - which is why that is a correct choice.

Which aspect are you confused by?

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##### Full Member
10+ Year Member
Okay so I think I remember this passage.

When the object drops from its original elevated position, it converts PE to KE. The amount of PE = mgh.

The amount of PE is also equal to the amount of KE the object gets after dropping.

To stop the object, friction needs to exert enough work to make that KE = 0.

Thus,

initial PE from elevated position = KE at bottom of ramp = work that friction must do.

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#### MelloTangelo

##### Full Member
10+ Year Member
Thanks for the help guys, and sorry for not including the page number! I'm still getting the hang of this, hah, but I'll be sure to do so next time.

I think my reading comprehension (or something along those lines) was failing for this passage, as I was looking at the movement from point b to c independently, and not accounting for anything before that point.

It makes complete sense, now, to look at the conversion of PE to KE, and the loss of that KE as the box comes to a halt.

Thanks again!
Seriously. <3

#### Willie_L88

##### New Member
7+ Year Member
Sorry to continue this thread. I understand the explanations given but why can't the answer be C?

Work = Fd and the question is asking for how much work is done by friction. Thus I put F to equal frictional force.

Ff = coefficient x Normal force = coefficient x mgcos(theta)

Work = coefficient x mgcos(theta) x change in distance from c-b

Therefore I put C. Why can't this also be the answer? Am I mixing up some theory or rule?

#### Cawolf

##### PGY-2
10+ Year Member
W = Fd = (mu)(Fn)d = (mu)mgd = (mu)mgd(c-b)

This is a valid answer and there are two reasons why it can't be C.

The normal force acts upwards (flat rollers) so there is no trig function AND there is no mention of the coefficient of friction.

The answer you wrote doesn't match C.

#### Willie_L88

##### New Member
7+ Year Member
Ok I understand now. Thank you