I was not there on thursday and did not have class today, but he told me to graph the data. It is better to be prepared and ahead of the game than not do it.
If you did finish the experiment, you should try to finish up both graphs, and if you have any questions or confusions, you'll be able to ask Mr. Crane tomorrow or get insight from the blog.
You definitely want to try and finish them on amp day so you can compare with others... If the daughter graph is linear write the equation of the line. Then Interpret the slope.
He also said that if we understood how, then we should try the daughter graph. Just like in that lab a few months ago. We have the example graph in our notebooks.
I don't think so, unless you have the units of the mins and maxes then you might be able to figure it out. If I were you I would just try to get more information as soon as possible.
I'm not satisfied with the results of my groups experiments... should I attempt to graph the daughter and see how it turns out, or should I come in and double check the experiment?
If you don't think that your data is correct, then you might want to redo the experiment just to be safe. The graph might turn out OK, but I honestly don't know. -Kevin Meglathery
Compare your results with another group in your class, and if your data is a little off, then it might be helpful to redo the experiment. You could try graphing the daughter graph, but like Kevin, I don't know how it would come out.
I'm confused... I tried graphing the daughter graph and it isn't completely linear.. Should I wait and talk to him or is it due in the beginning of class tomorrow? ~Katie Cabrera
Double check all of your points for the 1 over mass to see if it was a calculator error. If they're fine, then double check your graph's scale and that the coordinates are in the right spot.
Parent graph is mass on the x-axis and acceleration on the y-axis. Daughter graph is 1 over mass on the x-axis and accleration on the y-axis. -Kevin Meglathery
I'm pretty sure that we should get a positive slope. Because the daughter graph is 1 over the mass, smaller masses mean bigger fractions. (Would you rather have 1/300 of a dollar or 1/1550 of a dollar?) This means that they are farther from zero. -Kevin Meglathery
My b value was positive but I think its supposed to be 0 because a=F/m has a y-intercept of zero (no b).
We graphed the inverse of the mass(x) versus the acceleration(y). The graph was to be linear. The are described by y=mx+b. The line is a=m*(1/m)+b. We know a=F/m, or a=F*(1/m) so to get the acceleration you multiply our x, the inverse of the mass, by the force to get the acceleration. Thus, as we wrote a=m*(1/mass), the m=the applied force.
y=m*x+b a=m*(1/m) a=F*(1/m)
The force is the weight of the the hanging weight, which can be found using F=ma.
50 g (.05 kg) hanging weight, accelerating at 9.8 m/s/s, pulls the car with a force of .05*9.8= .49 N.
I don't think that you or anyone else could have explained that any better Billy. Now, if we can explain what the slope means to Mr. Crane, he said that he would cry. I kind of want to see that. -Kevin Meglathery
Billy, you didn't take into account the mass of the cart. I may be wrong but I thought the total mass included both the mass of the cart and the hanging weight. -Robyn K.
Yes. That is when the object reaches terminal velocity. It means that the force of gravity counteracts the force of air resistance. At that point, it won't accelerate any more. -Kevin Meglathery
An object in free fall falls at a constant speed, otherwise known as terminal velocity, when air resistance becomes as strong as gravity in the opposite direction. Thus, the net force becomes zero, and the object no longer accelerates.
The ice problem has odd wording for the answers, when you lay down, the total force doesn't change, but it rather spreads evenly throughout the area you are covering.
How negative was it? If it was "-.1", then I don't think it's a big deal. If it was "-10", then you might want to check your BLOF. It should be zero because if an object doesn't have mass, then it can't accelerate. -Kevin Meglathery
I don't think that's frowned upon, Noah, but the fundamental unit of force is Newtons. A pascal is probably related to a Newton because it measures a type of force, but a Newton isn't related to a pascal.
John, I believe you are almost correct. Because friction was small enough that we could ignore it, I think that acceleration acted the way it did because of inertia. Billy, you are right when F stays the same, which it did in our lab. Tom, it takes a larger force to move a larger object with the same acceleration. Remember that x=0 is an asymptote because no matter how much weight we put on the cart, it will still accelerate. -Kevin Meglathery
What equation should we use in order to figure out number 50 on the test, I am debating on what to use between; F=ma or Vf^2=Vo^2+2a(Change of P)since I believe both would help solve the problem, I believe, PLASE RESPONED.
Do whatever you need to do to solve the problem. What are you trying to find? How do you get what you are trying to find? You believe both would help. Use the equations you need to solve the problem.
You would end up using both. This test is based on Newton's 2nd Law, therefore, you're going to end up using F=ma. In order to use F=ma, however, you need the acceleration. Look at what information is given to you, choose the best equation, plug it in, and you should be good.
No matter how many N an object has, if you pull on it with a constant force, no matter how many Newtons you are pulling with, shouldn't the acceleration be 0?
The magnitude of a force is basically the amount of force applied, regardless of direction. You can infer this because the choices for those questions were measured in Newtons, which is the fundamental unit of force.
Retarding force is an object's resistance to motion. In this case, the retarding force is caused by the breaks of the car. Because this test is on the 2nd law, you most likely have to use F = ma. To find the acceleration, you would use the equation we received from Day S1 [(Final Velocity)^2 = (Initial Velocity)^2+2a(Change in Position) ]. You plug in chug to find acceleration, you already have the mass, and then you use F = ma to find the retarding force.
Not really. To get the answer, you still have to use F=ma, but it's just set up into a more complex situation. It could contain aspects of Newton's 3rd Law, though, because many real life instances do.
I did not finish both graphs during class. Should I do them for HW or will we have time on Monday?
ReplyDelete-Kevin Meglathery
I don't think you are alone, so hopefully we will have time on monday.
DeleteI was not there on thursday and did not have class today, but he told me to graph the data. It is better to be prepared and ahead of the game than not do it.
DeleteAgree w/ Mimi
DeleteI would do it for homework, since it would help us in class be ahead of schedule.
DeleteHarrison,
Deletesome kids need the tools in class to finish.
But yeah, you are right.
If you didn't finish the experiment, I would go in at lunch to check with Mr. Crane. There are graphs online too.
DeleteIf you did finish the experiment, you should try to finish up both graphs, and if you have any questions or confusions, you'll be able to ask Mr. Crane tomorrow or get insight from the blog.
DeleteYou definitely want to try and finish them on amp day so you can compare with others... If the daughter graph is linear write the equation of the line. Then Interpret the slope.
DeleteI talked to Mr. Crane. He told me to try to finish up the parent graph.
ReplyDeleteHe also said that if we understood how, then we should try the daughter graph. Just like in that lab a few months ago. We have the example graph in our notebooks.
ReplyDeleteI am a little confused of what we are graphing. Are we supposed to do the parent graph (curved) AND the linear graph?
ReplyDelete-Robyn K.
Yes, we need to do both.
Delete~Makena
Yes you need the Parent Grapg to determine how to make it lenear. See previous lab not as Easy as Pi.
ReplyDeleteMr. Crane will be out on Monday. Complete the practice Pages Newton' Second Law and the ones relating to the lab.
ReplyDeleteDo not do 3rd law.
should we do the pages in the packet in class tomorrow ?
DeleteI did the ones that talk about mass and weight, but I was kinda confused about Newtons second law stuff.
Delete~Makena
We have to complete the packet pages up to page 20.
Deletemy group only collected the slope on friday.Is that enough for the parent graph?
ReplyDeleteI don't think so, unless you have the units of the mins and maxes then you might be able to figure it out. If I were you I would just try to get more information as soon as possible.
DeleteI'm not satisfied with the results of my groups experiments... should I attempt to graph the daughter and see how it turns out, or should I come in and double check the experiment?
ReplyDelete~~Taylor Nardone
If you don't think that your data is correct, then you might want to redo the experiment just to be safe. The graph might turn out OK, but I honestly don't know.
Delete-Kevin Meglathery
Compare your results with another group in your class, and if your data is a little off, then it might be helpful to redo the experiment. You could try graphing the daughter graph, but like Kevin, I don't know how it would come out.
DeleteI'm confused... I tried graphing the daughter graph and it isn't completely linear.. Should I wait and talk to him or is it due in the beginning of class tomorrow? ~Katie Cabrera
ReplyDeleteIs only one point out of place? Or is the whole graph curved?
Delete-Kevin Meglathery
Double check all of your points for the 1 over mass to see if it was a calculator error. If they're fine, then double check your graph's scale and that the coordinates are in the right spot.
DeleteDo what Nazia said, and if it still isn't linear, then just ask Mr. Crane, show him the line and how you double checked it.
DeleteFor the parent graph, are we graphing the mass and acceleration?
ReplyDeleteand what are we graphing on the daughter graph?
Parent graph is mass on the x-axis and acceleration on the y-axis.
DeleteDaughter graph is 1 over mass on the x-axis and accleration on the y-axis.
-Kevin Meglathery
Did anybody else get a negative b value for the daughter graph?
ReplyDelete~Makena
I'm pretty sure that we should get a positive slope. Because the daughter graph is 1 over the mass, smaller masses mean bigger fractions. (Would you rather have 1/300 of a dollar or 1/1550 of a dollar?) This means that they are farther from zero.
Delete-Kevin Meglathery
The slope should be a postive number, or you did in your equations incorrectly
DeleteIf the slope was a negative number, that means you either wrote something down wrong, or you messed up all your equations.
Delete-Jake Naumann
People,
Deletey=mx+b
b=/=slope. b=y-intercept
m=slope
My b value was positive but I think its supposed to be 0 because a=F/m has a y-intercept of zero (no b).
We graphed the inverse of the mass(x) versus the acceleration(y). The graph was to be linear. The are described by y=mx+b. The line is a=m*(1/m)+b. We know a=F/m, or a=F*(1/m) so to get the acceleration you multiply our x, the inverse of the mass, by the force to get the acceleration. Thus, as we wrote a=m*(1/mass), the m=the applied force.
y=m*x+b
a=m*(1/m)
a=F*(1/m)
The force is the weight of the the hanging weight, which can be found using F=ma.
50 g (.05 kg) hanging weight, accelerating at 9.8 m/s/s, pulls the car with a force of .05*9.8= .49 N.
F=.49 N, m(slope)=.49
a=.49/m
I don't think that you or anyone else could have explained that any better Billy.
DeleteNow, if we can explain what the slope means to Mr. Crane, he said that he would cry. I kind of want to see that.
-Kevin Meglathery
Well he did say that slope =.49, and a=.48/m, so slope is acceleration?
Deleteacceleration=.49/mass
DeleteBilly, you didn't take into account the mass of the cart. I may be wrong but I thought the total mass included both the mass of the cart and the hanging weight.
Delete-Robyn K.
It does.
DeleteThe mass is the total mass of the system being accelerated.
I only calculated the force, which was only the hanging weight's mass*gravity, because that is what is pulling on the string that is pulling the cart.
So, is anyone else in the m=.47 club?
Delete-Kevin Meglathery
Right here, Kevin! When I told Mr. Crane my slope, he screamed pretty loudly and high pitched, so I would assume we're on the right track.
DeleteI got a=.248(1/m)+.203
Deletedefinitely did not measure that acceleration right
Your graph might be slightly off because your y-intercept should be 0 or as close to 0 as possible.
DeleteIn free fall, is it possible for an object to stay at a constant speed?
ReplyDeleteYes. That is when the object reaches terminal velocity. It means that the force of gravity counteracts the force of air resistance. At that point, it won't accelerate any more.
Delete-Kevin Meglathery
In class, Mr. Crane told us we would change velocity for 6 seconds in a free fall. After that we would be the same speed.
DeleteI agrees with your answer, Kevin, but I think you may have mix up the forces. Isn't it the air resistance counteracts the force of gravity?
DeleteIt may be. I really don't know. But in all honesty, does it really matter? The forces cancel, that's all I know.
Delete-Kevin Meglathery
An object in free fall falls at a constant speed, otherwise known as terminal velocity, when air resistance becomes as strong as gravity in the opposite direction. Thus, the net force becomes zero, and the object no longer accelerates.
DeleteThe ice problem has odd wording for the answers, when you lay down, the total force doesn't change, but it rather spreads evenly throughout the area you are covering.
ReplyDeleteFor my b value I had gotten a negative answer is this correct or could someone steer me on the right direction to get the right answer.
ReplyDeleteHow negative was it? If it was "-.1", then I don't think it's a big deal. If it was "-10", then you might want to check your BLOF.
DeleteIt should be zero because if an object doesn't have mass, then it can't accelerate.
-Kevin Meglathery
We never went over the term pascal in class, right?
ReplyDeleteright
Deleteidk whether or not this is frowned upon, but i googled it and it is a unit of pressure and pressure is a force, so i assume that it is a unit of force
DeleteI don't think that's frowned upon, Noah, but the fundamental unit of force is Newtons. A pascal is probably related to a Newton because it measures a type of force, but a Newton isn't related to a pascal.
DeleteI'm a little confused. In our previous lab, is the reason that acceleration varies inversely with mass because of friction?
ReplyDeleteNo. Acceleration is always varies inversely with mass because a=F/m.
DeleteI don't think so, I think it just takes larger force to move a larger object.
DeleteJohn, I believe you are almost correct. Because friction was small enough that we could ignore it, I think that acceleration acted the way it did because of inertia.
DeleteBilly, you are right when F stays the same, which it did in our lab.
Tom, it takes a larger force to move a larger object with the same acceleration. Remember that x=0 is an asymptote because no matter how much weight we put on the cart, it will still accelerate.
-Kevin Meglathery
What equation should we use in order to figure out number 50 on the test, I am debating on what to use between; F=ma or Vf^2=Vo^2+2a(Change of P)since I believe both would help solve the problem, I believe, PLASE RESPONED.
ReplyDeleteDo whatever you need to do to solve the problem. What are you trying to find? How do you get what you are trying to find? You believe both would help. Use the equations you need to solve the problem.
DeleteYou would end up using both. This test is based on Newton's 2nd Law, therefore, you're going to end up using F=ma. In order to use F=ma, however, you need the acceleration. Look at what information is given to you, choose the best equation, plug it in, and you should be good.
DeleteThat was a great explanation. Thanks.
DeleteYou guys just saved my life. Thanks!
DeleteMike Keough
No matter how many N an object has, if you pull on it with a constant force, no matter how many Newtons you are pulling with, shouldn't the acceleration be 0?
ReplyDeleteAn unbalanced force will make an object accelerate (sum of forces *net force* =/= 0).
DeleteBalanced forces result in an object not accelerating (not changing its velocity; sum of forces *net force* =zero).
a=F/m
An object's mass (in kg), not its own force (in N) affect its acceleration.
The object would accelerate right when you start to pull on it, but if you maintain a constant force, then yes, a=0.
Delete-Kevin Meglathery
This comment has been removed by the author.
ReplyDeleteCan someone explain to me what "magnitude of the force" means?
ReplyDelete-Jake Naumann
The magnitude of a force is basically the amount of force applied, regardless of direction. You can infer this because the choices for those questions were measured in Newtons, which is the fundamental unit of force.
DeleteDo they still cancel out when directly opposite?
DeleteI would assume that they would, but you might want to double check with someone else on that one.
DeleteCould someone please explain what retarding force s and what equation you would use to solve for it.
ReplyDeleteRetarding force is an object's resistance to motion. In this case, the retarding force is caused by the breaks of the car. Because this test is on the 2nd law, you most likely have to use F = ma. To find the acceleration, you would use the equation we received from Day S1 [(Final Velocity)^2 = (Initial Velocity)^2+2a(Change in Position) ]. You plug in chug to find acceleration, you already have the mass, and then you use F = ma to find the retarding force.
Deletebut for this question we are already given the acceleration, so if I do mass times acceleration = the force will that just be my retarding force.
Deletenever mind I thought it was acceleration it is the velocity my bad.
DeleteIt's alright. You had the general understanding of it, and if the acceleration was given, that's all you would have had to do have done.
DeleteWhen converting, 1 kg is equal to about 9.8 newton's right?
ReplyDeleteYes.
Deletethank you
DeleteCan someone explain to me what a net force is exactky?
ReplyDeleteIt's simply the total or sum of the forces. You just add everything together, but don't forget to consider their direction.
DeleteNet Force=Total Force
ReplyDeleteDoesn't question 53) require Newton's 3rd law?
ReplyDeleteNot really. To get the answer, you still have to use F=ma, but it's just set up into a more complex situation. It could contain aspects of Newton's 3rd Law, though, because many real life instances do.
DeleteTo find 52), would the tension in the string be the same as the force pulling the 8.8kg block?
Delete