I used the bottom of the circle as point Q since like Cleo says, it didn't specify. I figured it would be easier to calculate the Kinetic Energy at the bottom.
for the 19993 B1 elevator problem, the only forces I have acting on the kid are FN and Fg. To solve for the acceleration during each time interval, I just assumed FN remained the same, which I found to be -Fg, since the sum of the fores in the y is equal to fg and fn. So, I got Fn equals 500 N, and then I just solved for all of the varying accelerations using FN equals 500 N and the varying Fgs. Does anyone else agree with that because I'm not 100% sure?
I have not found out what it is, but there must be another force acting on the person in order to have him move with the elevator. That's what I was thinking
Because the elevator is moving wouldn't Fg+Fn=ma not zero because Fn would change with the movement of the elevator and Fg would go unchanged because mass and gravity never change?
I rearranged F=ma to a=F/m to solve for this one. I used the force in newtons that was on the graph to plug in for each of the time intervals. And I used Fg that was given to find the m (which was 50kg). For instance, for the time interval 5-10, the F changed from 500 to 700. Which means the delta F is 200. So to get acceleration, I plugged in a=200/50 to get 4m/s^2. I don't know if that's correct but it kind of makes sense? Maybe?
for question #2b it says momentum. are we supposed to look up the formulas for that or is there a way to solve that problem with what we have learned so far.
I think the velocity of the sled is the same as the horizontal velocity of the block because horizontal velocity is constant and the sled starts at rest. The equation for momentum is mass x velocity, so I think the momentum would be (M1+M2) x Vo... hope this helps
I agree with Morgan that the horizontal velocity is a constant. I assumed the PE before was equal to the PE after. I set mv = (M1+M2) vo and solved for velocity to get my answer.
For #1a ( on the third page ) are the only forces acting on the object at the top of the circle Ft and Fg? Maybe I don't have the right idea of a vertical circle, but I don't think there would be friction...
For the picture on this problem (#1) did you draw the forces on the box as if it were at the bottom of the path even though it states those are the conditions at the top of path
I said that that there would be Ft and Fg acting on it at the top of circle. And I used the bottom of the circle as point Q because I figured it would be easier to solve for the difference in Kinetic Energy that way.
I solved for the coefficient of friction...but instead of having an equals sign I tried to determine if it needed to be greater than or less than the other side of the equation
I did something similar to what Caity did. So basically, I used the variables given to determine the relationship between the coefficient of friction and v,r, and g and then I determined where the coefficient of friction would be less than, greater than or equal to that relationship.
I think you use Q=(m)(c)(delta T) to solve it and you are solving for delta T. So I rearranged it to delta T=Q/mc. Q is the energy which you solved for in part c so it should just be plug and chug. I don't know if that makes any sense.
How do you go about solving for a and b of this problem? For a I got the square root of (Fnx*r/m) and b I got M*2mg but that doesnt seem right. Maybe I am not seeing it, but how do you start to solve it?
Morgan- during class when we were going over work we said that velocity is the square root of 2gh, so I used that thought for part a.
For part b, I said that the PE:initial = PE:final. Then plugged in for the PE, cancelled the masses, and substituted the answer to part a for the initial velocity.
For problem B1 are people getting numbers as answers? for part b i understood that we just needed an equation, but for c how do we solve for the final distance without knowing the time?
For B i got the answer in terms of time. it was an equation. I don't know about C. I used "xf=.5at^2 + vot + xi" so I have it all in terms of t right now. I don't know where we would get a time to plug in though.
I solved for part a of B1 using energy. I used the formula for kinetic energy since the block was in motion. I modeled my equation off of vf-vi, only I used the KEf- KEi = the work done. So, w= 1/2mvf^2- 1/2mvi^2. We are given the initial velocity and the final velocity would be 0, so after plugging everything in, I got the work done to be 72 joules. That's just how I did it, but I'm not 100% certain if it is correct.
I got the same answer using W=mad... mine is -72 because it is the work to bring it to rest, but the sign depends on frame of reference. So i think you're right, Cleo
1995B3 part e Is the tension of this system the same as if it's at rest? I'm sort of confused whether the ball has its on Fc, if its tension affected by the Fc of the cart, or if its not affected at all.
for e, there are two parts. Th is zero because there is no horizontal acceleration which means there is no horizontal tension. for Tv,you would use mv^2/r=mg-Tv. That is what i did to find each of those tensions.
For number 1, (not B1) where exactly is point Q? Are we supposed to find a point a label it Q?
ReplyDeleteI'm just assuming that point Q is any random point on the circle since it doesn't specify where point Q is.
DeleteI used the bottom of the circle as point Q since like Cleo says, it didn't specify. I figured it would be easier to calculate the Kinetic Energy at the bottom.
Deletefor the 19993 B1 elevator problem, the only forces I have acting on the kid are FN and Fg. To solve for the acceleration during each time interval, I just assumed FN remained the same, which I found to be -Fg, since the sum of the fores in the y is equal to fg and fn. So, I got Fn equals 500 N, and then I just solved for all of the varying accelerations using FN equals 500 N and the varying Fgs. Does anyone else agree with that because I'm not 100% sure?
ReplyDeleteI have not found out what it is, but there must be another force acting on the person in order to have him move with the elevator. That's what I was thinking
DeleteI think it is just Fg and Fn... I solved it how you did, Cleo
DeleteBecause the elevator is moving wouldn't Fg+Fn=ma not zero because Fn would change with the movement of the elevator and Fg would go unchanged because mass and gravity never change?
DeleteI rearranged F=ma to a=F/m to solve for this one. I used the force in newtons that was on the graph to plug in for each of the time intervals. And I used Fg that was given to find the m (which was 50kg). For instance, for the time interval 5-10, the F changed from 500 to 700. Which means the delta F is 200. So to get acceleration, I plugged in a=200/50 to get 4m/s^2. I don't know if that's correct but it kind of makes sense? Maybe?
DeleteAleena- that is what I got also.
Deletefor question #2b it says momentum. are we supposed to look up the formulas for that or is there a way to solve that problem with what we have learned so far.
ReplyDeleteI think the velocity of the sled is the same as the horizontal velocity of the block because horizontal velocity is constant and the sled starts at rest. The equation for momentum is mass x velocity, so I think the momentum would be (M1+M2) x Vo... hope this helps
DeleteI agree with Morgan that the horizontal velocity is a constant. I assumed the PE before was equal to the PE after. I set mv = (M1+M2) vo and solved for velocity to get my answer.
DeleteFor #1a ( on the third page ) are the only forces acting on the object at the top of the circle Ft and Fg? Maybe I don't have the right idea of a vertical circle, but I don't think there would be friction...
ReplyDeleteYeah that is what I had for the forces as well and there cant be friction because nothing is rubbing against the object.
DeleteFor the picture on this problem (#1) did you draw the forces on the box as if it were at the bottom of the path even though it states those are the conditions at the top of path
DeleteAlso where is point q in the problem?
DeleteI said that that there would be Ft and Fg acting on it at the top of circle. And I used the bottom of the circle as point Q because I figured it would be easier to solve for the difference in Kinetic Energy that way.
DeleteJared- I drew it with Ft and Fg acting in the same direction. It would be similar to the FBD at the bottom, but at the top tension works with gravity
DeleteI looked up how the picture was supposed to look and point P was at the top of the circle and point Q was at the opposite end, on the bottom.
DeleteFor B2(b) [on the second page] what are we trying to find exactly?
ReplyDeleteYou are solving for the combined velocity of the toboggan and the block using the momentum equation.
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DeleteI solved for the coefficient of friction...but instead of having an equals sign I tried to determine if it needed to be greater than or less than the other side of the equation
DeleteI did something similar to what Caity did. So basically, I used the variables given to determine the relationship between the coefficient of friction and v,r, and g and then I determined where the coefficient of friction would be less than, greater than or equal to that relationship.
DeleteFor problem1994b2 how do we solve for specific heat and thermal energy using those variables?
ReplyDeleteI think you use Q=(m)(c)(delta T) to solve it and you are solving for delta T. So I rearranged it to delta T=Q/mc. Q is the energy which you solved for in part c so it should just be plug and chug. I don't know if that makes any sense.
DeleteHow do you go about solving for a and b of this problem? For a I got the square root of (Fnx*r/m) and b I got M*2mg but that doesnt seem right. Maybe I am not seeing it, but how do you start to solve it?
DeleteMorgan- during class when we were going over work we said that velocity is the square root of 2gh, so I used that thought for part a.
DeleteFor part b, I said that the PE:initial = PE:final. Then plugged in for the PE, cancelled the masses, and substituted the answer to part a for the initial velocity.
For problem B1 are people getting numbers as answers? for part b i understood that we just needed an equation, but for c how do we solve for the final distance without knowing the time?
ReplyDeleteFrom what I read, I think we are getting formulas for each answer according to what is being asked; not numbers.
DeleteFor B i got the answer in terms of time. it was an equation. I don't know about C. I used "xf=.5at^2 + vot + xi" so I have it all in terms of t right now. I don't know where we would get a time to plug in though.
DeleteI solved c by using the equation Vf^2=Vo^2+2ad... hopefully this helps
Deletefor this one if your using "xf=.5at^2 + vot + xi" as the equation. would you use the quadratic formula to find t?
DeleteI solved for part a of B1 using energy. I used the formula for kinetic energy since the block was in motion. I modeled my equation off of vf-vi, only I used the KEf- KEi = the work done. So, w= 1/2mvf^2- 1/2mvi^2. We are given the initial velocity and the final velocity would be 0, so after plugging everything in, I got the work done to be 72 joules. That's just how I did it, but I'm not 100% certain if it is correct.
ReplyDeleteI got the same answer using W=mad... mine is -72 because it is the work to bring it to rest, but the sign depends on frame of reference. So i think you're right, Cleo
DeleteI got the same thing using Cleo's way. Mine is negative, like Morgans...so I'm assuming that's still ok due to whatever frame of reference we picked?
DeleteYeah, I got negative too,but I didn't know if work could be negative so I just assumed that the negative meant that's how much energy was lost.
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ReplyDelete1995B3 part e Is the tension of this system the same as if it's at rest? I'm sort of confused whether the ball has its on Fc, if its tension affected by the Fc of the cart, or if its not affected at all.
ReplyDeletefor e, there are two parts. Th is zero because there is no horizontal acceleration which means there is no horizontal tension. for Tv,you would use mv^2/r=mg-Tv. That is what i did to find each of those tensions.
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