lnphan study questions

[blue0rchid]

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Hey guys, I'm gonna put all my questions here. Whether they can be answer or not, I'll really appreciate all the help and support given from this forum.

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Physics - Friction

Question: The work done on a road surface, as a car skids across it, is:
A - negative & directly proportional to the skidding distance
B - negative & inversely proportional ...
C - positive & directly proportional ...
D - positive & inversely proportional ...

Btw, this is BR example 3.2b. My choice would be either A or C. I'm confused a bit about work done on an object. If the work is done by friction, on the car, then it's definitely negative & proportional to skidding distance. But what about work done on the road? Wouldn't it be the opposite? But the road doesn't move... On second thought, initial energy of the road is zero, final energy is positive because of heat transfer (right?). Initial energy of the car is positive, final energy is zero, so its kinetic energy was transferred somewhere... So, answer should be C ? *Sigh*

Still don't have an answer for this problem yet.

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Physics - EK 1001 #430 - SOLVED - Thanks justhanging

Question: I drop 1kg lump of clay at 1m high on top of another 1kg lump of clay sitting in the mud. The clumps of clay stick together and driving to the mud a distance of 1cm. What's the average force exerted on the mud?
A - 500N
B - 520N
C - 1000N
D - 1020N

My solution: I found v0 of clump1 right before it hits clump2: v0 = sqrt(2gh) with h=1m
Since clump1 and clump2 have the same mass, they will both have velocity after impact at v1=(v0)/2
Call v2=0 the final velocity after both clumps travel 1cm into the mud (and stop).
I have (v2)^2 = 0 = (v1)^2 + 2*a*x . F=ma so a=F/m with m being the mass of both clumps (m=2kg) and x=1cm=0.01m
So, I have 0 = 2gh/4 + 2*(F/m)*x ---> F=500N

I chose A as my answer, but EK said it's wrong. Correct answer is B.
According to EK, they use work/energy formulas to calculate F (because they treat the mud like friction):
F d = 1/2 m (v1)^2 + m g h with d=h=0.01m. And they come out with 520N.

Can someone point out what's wrong w/ my solution? What is it that I am missing? What's the flaw in my logic for getting 500N instead of 520?
Answer is in http://forums.studentdoctor.net/showpost.php?p=11034377&postcount=6

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Physics - Gravitational force - SOLVED - Thank you.

It's the BR example 2.5b. A satelite orbits around Venus in eliptical orbital shape, moving counter clock-wise from A to D. Point A is to the right of Venus and furthest from Venus, point B is right on top Venus (a bit north west), point C is West of Venus and closet to Venus. Point D is South of Venus and kinda in between point A and C.

Question: At which point in its orbit is satelite moving the fastest
Choices: A , B , C , Its speed is constant

I'm thinking point C. Because at point A (closest to Venus) it has highest acceleration. What do you think?

I found the answer, with Kepler's Laws or Orbital Motion (second law). Point C is the right answer.

 

[GreenRabbit]

Conservation of energy still applies. The furthest point has the greatest potential energy, and as it orbits to the closest point that potential energy is converted into kinetic energy. As it orbits away from C back to A, the kinetic energy reverts to potential energy. In other words, we know kinetic energy is greatest at the closest point without considering acceleration.

 

[justhanging]

You can use centripical force to solve it as well. When the satelite is at its closets position it has the greatest force due to gravity but this force is the centripical force which is equal to mv^2/r.

 

[blue0rchid]

Urgh, can someone help me answer my question about EK Physics 1001, #430?
Thanks.

 

[justhanging]

I believe the force you found was the net force, which includes friction of the mud pointing up minus gravity pointing down. The question was just asking for the mud force, that is why your answer is a bit less. What is the force of gravity? 2mg = 20 and that is where the missing 20 comes from.

 

[blue0rchid]

I believe the force you found was the net force, which includes friction of the mud pointing up minus gravity pointing down. The question was just asking for the mud force, that is why your answer is a bit less. What is the force of gravity? 2mg = 20 and that is where the missing 20 comes from.

I see... Thank you.

So imagine the 2 clumps have gravitational force on the mud Fg = 20N (which is normally canceled out by normal force). Apply another F=500N to drive the clumps in. So total force on the mud is 520N.

 

[justhanging]

No the frictional force is 520. The total force is the net force which is 500, what you calculated.

500 = 520 - 20

hopefully that clears it up.

 

[Catburr]

Question: The work done on a road surface, as a car skids across it, is:
A - negative & directly proportional to the skidding distance
B - negative & inversely proportional ...
C - positive & directly proportional ...
D - positive & inversely proportional ...

Btw, this is BR example 3.2b.

Car is doing the work. Car is losing energy, since it's slowing down. See the rules just before Ex 3.1, "If positive work is done by an object, the object loses energy." So the car must be doing positive work on the road. Work is force x distance, so yup, the work is directly proportional to distance. It's C.

Also, I'll mention because I didn't figure this out until a couple chapters into BR myself, the answers to all the x.xb problems are in the back, on page 282, after the index. I think there's a BR physics newer than mine (which has a blue cover, copyright 2009), but hopefully the new version has the answers too. No explanations unfortunately.