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The wax wears off so that the coefficient of kinetic friction is now $\mu_{k}$. How much force does the sail now have to provide for a constant velocity ride back up the hill (once it gets started)?A. $\mu_{k} g \sin \alpha$B. $\mu_{k} w \sin \alpha$C. $\mu_{k} w \cos \alpha$D. $w \sin \alpha+\mu_{k} w \cos \alpha$E. $w \sin \alpha-\mu_{k} w \cos \alpha$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

Lectures

04:01

2D kinematics is the study…

03:28

Newton's Laws of Moti…

01:48

If the toboggan is well wa…

05:01

A 16 -kg sled starts up a …

01:56

A horse is trotting along …

03:20

A sled of mass $m$ is give…

04:39

mmh A toboggan slides down…

02:20

$\mathrm{A} 1000 \mathrm{k…

03:53

An $85-\mathrm{kg}$ skier …

05:40

A 60.0 -kg person, running…

01:49

A 16-kg sled is being pull…

01:14

A girl exerts a $36-\mathr…

00:42

A sled of mass $5.0 \mathr…

01:10

An iccboat is at rest on a…

08:40

Sam, whose mass is $75 \ma…

01:21

A child sleds down a $12^{…

05:58

A skier with a mass of $63…

09:01

A bicyclist of mass 75 $\m…

01:05

After the skier on the $37…

04:04

(III) A bicyclist can coas…

03:57

(II) The coefficient of ki…

0:00

(II) A skier moves down a …

anyone deals with some of the examples presented earlier in the chapter. Examples 5.65 point 11 5 12 which is with the bobbin sliding down a hill. Ah, in the scenario presented for these last three problems of the chapter, the bargain has a sale added to it such that instead of traveling down the hill, do you want to travel up the hill on the hill slope? Thank Alfa And the weight of the toboggan, including people inside, is given by W. So these are the scenarios drawn here. All of these values are figures else Able came from example 5 12 on page 1 43 So I didn't derive any of these. So this is just my starting point. The question Anyone states that the wax wears off so that the coefficient of kinetic friction is now I'm UK. How much force does a sale now have to provide for a constant velocity ride back up the hill? I mean, you look at the components acting in our here in X direction because it's angled along the slope of the incline. So instead of traveling down me to travel up the hill in this direction to the net forces in the X direction here would be UK forces connect friction times the normal force which, if you look at here, has to be balanced by, um, the wait times coastline Alfa That's the expression for the normal minus r w sign Alfa Just looking at the net expression in terms off, uh, each component sooner or overcome this, then they must be acting. Um, I guess originally, this was I drew this in the downward scenario where my forces can expression would be traveling up to the left. But now that's not true. How Jordan red So this is now for traveling up the ramp in the forsaken activation will also be acting in the same directions. This is F k. So that scenario, they're both me say they're both positive because they're contributing in the same direction. So it would be UK time W Coast data plus w signed data. So this is the net force acting in the extraction which corresponds to a value provided an answer D given in our textbook, they go

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