TBR Physics Chapter 2 passage 3 question 20

[Dipraman]

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The question says, "A skier is given a strong push so that he slides up the hill for a certain distance (coefficient of kinetic friction is 0.01). When he gets to the highest point, he slides back down. How does the acceleration of the skier on his ascent compare to the acceleration on his descent? (Do not consider the acceleration during the initial push)

A. The acceleration on the descent is smaller in magnitude than on the ascent
B. The acceleration on the ascent is smaller in magnitude than on the descent
C. Both accelerations are same
D. The accelerations have the same magnitude but different directions.

I put B because during the ascent, you will have the gravitational force (x component of weight) and force due to friction both act in the same direction to slow you down. Hence slower acceleration as compared to your descent because the gravitational pull will be in your favor. But the book says it A 😕 Can anyone please help me understand this?

 

[dudewheresmymd]

The question says, "A skier is given a strong push so that he slides up the hill for a certain distance (coefficient of kinetic friction is 0.01). When he gets to the highest point, he slides back down. How does the acceleration of the skier on his ascent compare to the acceleration on his descent? (Do not consider the acceleration during the initial push)

A. The acceleration on the descent is smaller in magnitude than on the ascent
B. The acceleration on the ascent is smaller in magnitude than on the descent
C. Both accelerations are same
D. The accelerations have the same magnitude but different directions.

I put B because during the ascent, you will have the gravitational force (x component of weight) and force due to friction both act in the same direction to slow you down. Hence slower acceleration as compared to your descent because the gravitational pull will be in your favor. But the book says it A 😕 Can anyone please help me understand this?

On his ascent, the forces acting on the skier are mgsintheta paralell to the slope towards the bottom of the slope and friction in the same direction as mgsintheta. On the descent, mgsintheta is parallel to and towards the bottom of the slope, but now friction is pointed towards the top of the slope in the direction opposite to mgsintheta.

Sum Forces=ma

Therefore, on the ascent the acceleration's magnitude will be larger b/c the two forces are in the same direction as opposed to on the descent where they oppose one another...force is proportional to acceleration:

Fascent=mgsintheta+force due to friction
Fdescent=mgsintheta-force due to friction

They are asking for the magnitude of the acceleration. Thus, because Fascent>Fdescent in magnitude, acceleration ascent > acceleration descent in magnitude (F=ma).

 

[Dipraman]

Ah gotcha! Thanks

I guess I was just thinking about acceleration exclusively during ascent and descent. I guess during the ascent it should technically be "deceleration"

 

[zhealthcare]

On his ascent, the forces acting on the skier are mgsintheta paralell to the slope towards the bottom of the slope and friction in the same direction as mgsintheta. On the descent, mgsintheta is parallel to and towards the bottom of the slope, but now friction is pointed towards the top of the slope in the direction opposite to mgsintheta.

Sum Forces=ma

Therefore, on the ascent the acceleration's magnitude will be larger b/c the two forces are in the same direction as opposed to on the descent where they oppose one another...force is proportional to acceleration:

Fascent=mgsintheta+force due to friction
Fdescent=mgsintheta-force due to friction

They are asking for the magnitude of the acceleration. Thus, because Fascent>Fdescent in magnitude, acceleration ascent > acceleration descent in magnitude (F=ma).

This may seem dumb, but what happened to the initial force that was used to push the skier up the hill? Did we ignore it in our calculations? When I did the free body diagram for the skier on his ascent, I had the initial push force uphill (the only one in that direction) and the component of gravity (mg Sin theta) and the kinetic frictional force downhill. So uphill i ended up with the following:
Fnet = ma
Fpush - Fk - mgSintheta = ma
What am i doing wrong. Plz help

 

[FeinMS]

This may seem dumb, but what happened to the initial force that was used to push the skier up the hill? Did we ignore it in our calculations? When I did the free body diagram for the skier on his ascent, I had the initial push force uphill (the only one in that direction) and the component of gravity (mg Sin theta) and the kinetic frictional force downhill. So uphill i ended up with the following:
Fnet = ma
Fpush - Fk - mgSintheta = ma
What am i doing wrong. Plz help

That force is a "one-time" force that doesn't apply continuously during ascent or descent. You can disregard it.

 

[zhealthcare]

Thanks much

 

[ReadandPlay]

I'm trying to understand this!

Does kinetic friction always opposes motion? What about when you are ice skating does the friction add to you sliding more (motion)?

Also based, kinetic friction opposes motion so it points in the direction opposite to the direction of movement.
When the skier is going up the hill (ascending) there are two forces pointing in the same direction but his motion is opposing both friction and gravity, but when he is descending down the hill mg is same direction as his motion and friction is in the opposite direction, therefore he is only opposing one of the forces so his acceleration should be greater when he is descending then?

 

[techfan]

Yes, his acceleration would be greater during descent if you include direction, but this is asking for magnitude. It's hard to perform a thought experiment on this one since it is hard to imagine a force being applied (push force) that doesn't factor into changing the motion. The skier's pushed force is countered by both gravity and friction acting in the same direction (down the slope), but now pretend that the pushed force doesn't exist (set it equal to 0) and you'll get friction and gravity negative, but the magnitude makes it positive. Whereas when the skier is sliding down gravity is opposite to the friction force and you subtract the two and get a smaller magnitude.

 

[ReadandPlay]

Oh I get it now. Thanks again!