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Physics Questions

AgaBoogaBoo

AgaBoogaBoo

Lifer
I'm lost on these problems, not really looking for the answer because I have that, more for how to get started. When I say get started, that is right after identifying what was given to us.

First problem:
A force of 20N accelerates a 9kg wagon at 2.0m/s^2 along the sidewalk. How large is the frictional force and what is the coefficient of friction?

No clue about how large it is, but for the frictional force, we need the force due to kinetic friction and the normal force. I know the kinetic friction is 20N, right? How do I do the rest?

Second Problem
A 2kg brick has a sliding coefficient of friction of .38. What force must be applied to the brick for it to move at a constant velocity?

I think like the first problem, I need to apply 2kg somehow, not sure how though
 
Originally posted by: AgaBoogaBoo
I'm lost on these problems, not really looking for the answer because I have that, more for how to get started. When I say get started, that is right after identifying what was given to us.
Click to expand...

1. Draw a free body diagram.
2. Think physics
 
Originally posted by: AgaBoogaBoo
I'm lost on these problems, not really looking for the answer because I have that, more for how to get started. When I say get started, that is right after identifying what was given to us.

First problem:
A force of 20N accelerates a 9kg wagon at 2.0m/s^2 along the sidewalk. How large is the frictional force and what is the coefficient of friction?

No clue about how large it is, but for the frictional force, we need the force due to kinetic friction and the normal force. I know the kinetic friction is 20N, right? How do I do the rest?

Second Problem
A 2kg brick has a sliding coefficient of friction of .38. What force must be applied to the brick for it to move at a constant velocity?

I think like the first problem, I need to apply 2kg somehow, not sure how though
Click to expand...

Ff = coefficient * normal force
normal force = mass*gravity
So for problem 2, in order for the sum of the forces = 0 (constant velocity) find the frictional force. Use this information to do problem 1.
 
Originally posted by: Legendary
Originally posted by: AgaBoogaBoo
I'm lost on these problems, not really looking for the answer because I have that, more for how to get started. When I say get started, that is right after identifying what was given to us.

First problem:
A force of 20N accelerates a 9kg wagon at 2.0m/s^2 along the sidewalk. How large is the frictional force and what is the coefficient of friction?

No clue about how large it is, but for the frictional force, we need the force due to kinetic friction and the normal force. I know the kinetic friction is 20N, right? How do I do the rest?

Second Problem
A 2kg brick has a sliding coefficient of friction of .38. What force must be applied to the brick for it to move at a constant velocity?

I think like the first problem, I need to apply 2kg somehow, not sure how though
Click to expand...

Ff = coefficient * normal force
normal force = mass*gravity
So for problem 2, in order for the sum of the forces = 0 (constant velocity) find the frictional force. Use this information to do problem 1.
Click to expand...

Thanks 🙂

I think my teacher left that out of the notes for our class somehow 🙁
 
First problem:
if there were no friction, a force of 20N (potential) would turn it to kinetic energy of 0.5*m*v^2 in its entirety. however, this is not the case... so you can get the energy spent on battling the friction by calculating 20N - 0.5*9kg*2.0m/s^2 = 2N. Frictional force is given by frictional coefficient * normal force. since the normal force here would be 9kg*9.81N/kg, you can derive the coefficient by 2/(9*9.81)=0.022652622041001245894212255068524
i might be wrong tho 😛
 
Ok, the equation to use is this:

Fk = MkFn

which comes out to be:

20N = (Mk)(90)
Friction = 0.22

btw, Mk = friction
 
Second problem : to maintain inertia, all outside forces must sum to a zero. this means frictional force = - force applied
 
Originally posted by: AgaBoogaBoo
I'm lost on these problems, not really looking for the answer because I have that, more for how to get started. When I say get started, that is right after identifying what was given to us.

First problem:
A force of 20N accelerates a 9kg wagon at 2.0m/s^2 along the sidewalk. How large is the frictional force and what is the coefficient of friction?

No clue about how large it is, but for the frictional force, we need the force due to kinetic friction and the normal force. I know the kinetic friction is 20N, right? How do I do the rest?

Second Problem
A 2kg brick has a sliding coefficient of friction of .38. What force must be applied to the brick for it to move at a constant velocity?

I think like the first problem, I need to apply 2kg somehow, not sure how though
Click to expand...

Part A) You have two forces, the 20N force pushing it forward and the frictional force pushing in the opposite direction. You have the mass and the acceleration so you have net force. QED... you're almost done.

Part B) To move at a constant velocity, you need a net force of 0. You'll again have two forces, one that you apply and the other from the frictional force going in the opposite direction. QED... you're almost done.
 
if you need help with physics, ask me...i'm doing pretty well, or ask DrPizza...he's a physics teacher 😉
 
Originally posted by: VanillaH
lets check our answers... mine : 7.4556N?
Click to expand...

i got that as the counter force...i thought i needed to do something with the action force as well, no?
 
Originally posted by: TuxDave
Originally posted by: VanillaH
good, now help with mine 😛
Click to expand...

Step 1) F=MA
Step 2) The rest is obvious.... 😉
Click to expand...

no, mine is in another thread... its about p2p file sharing 😀

i got that as the counter force...i thought i needed to do something with the action force as well, no?
Click to expand...
i wouldnt think so... maybe you are reading into it too much 🙂
 
I'd have broken the first problem down into simple pieces, as a few others have done. 9kg accelerates at 2m/s/s Using Fnet=ma, the net force must be 9kg * 2m/s/s = 18N. Since there's an applied force of 20N, the frictional force must be 2N. The equation relating the coefficient of friction has already been given a couple of times above. To calculate the normal force, you can use F=mg. However, this is only true if the sidewalk is level, which I assume everyone is assuming. (additionally, it messes up the other calculations as well if the sidewalk is not level) I have a paper to write tonight, otherwise I'd type up a general solution for the case of a non-level sidewalk. It's always fun to mess with the teachers/profs that way.


Actually, I have a question for anyone with specific knowledge on this: is the coefficient of rolling friction calculated the same way as the coefficient of static or kinetic friction? I noticed the OP considered this a kinetic friction problem. That's a common misconception. When a tire is rolling on the ground, the two surfaces are not moving with respect to each other. Kinetic friction would be if the wheels were locked up and the tires were skidding, rather than just rolling. The frictional effects in the case of a rolling tire are primarily due to the changing deformation of the tires (hence an over-inflated tire rolls easier since it deforms less) as well as friction in the bearings. However, as I think about it, it seems that friction on the bearings could occur in a horizontal direction and be independent of the mass of the wagon. e.g. the coefficient of friction between brake pads of a car and the rotors have little to do with the mass of a car.
 
Originally posted by: DrPizza
Actually, I have a question for anyone with specific knowledge on this: is the coefficient of rolling friction calculated the same way as the coefficient of static or kinetic friction? I noticed the OP considered this a kinetic friction problem. That's a common misconception. When a tire is rolling on the ground, the two surfaces are not moving with respect to each other. Kinetic friction would be if the wheels were locked up and the tires were skidding, rather than just rolling. The frictional effects in the case of a rolling tire are primarily due to the changing deformation of the tires (hence an over-inflated tire rolls easier since it deforms less) as well as friction in the bearings. However, as I think about it, it seems that friction on the bearings could occur in a horizontal direction and be independent of the mass of the wagon. e.g. the coefficient of friction between brake pads of a car and the rotors have little to do with the mass of a car.
Click to expand...
As long as the wheels roll without slipping, it's considered static friction, and so is calculated in the standard way.
 
I just searched through a couple of college physics books. I couldn't find anything about rolling friction. But, now that I've thought about it, in problem 1, there is no coefficient of friction to calculate. The problem is too complex - it depends on where the friction comes from. Imagine that the tires are held onto the wagon with bolts and washers. If you tightened the bolts too much, it's going to create friction when you attempt to spin the wheels. The amount of friction between the washer/wheel/wagon is independent of the mass of the wagon (and therefore the normal force).

Pour some mass into your wagon - enough to double the mass, and I highly doubt you'll double the force of friction. The force will be somewhere between the original force of friction and twice the original force. I'm on a physics listserv - I'll ask that question of all the other physics teachers/profs on the same listserv tomorrow.
 
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