IB Physics I

  • #1
Hi, I'm in IB physics and I'm having a little trouble since the teacher would rather play computer games than teach us (the book isn't great either). Right now we're working on force and I was wondering if anyone could clarify a few things for me please.I'll just work through some problems and it would be great if someone could tell me if I'm wrong. Thanks!

So, basically, the most important thing in dealing with force is: F(net)= F(app)- F(res).

1. A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 seconds. The car is uniformly accelerated during the entire time. What net force is exerted on it?

So, acceleration = F(net)/m

V= d/t ; v = 40.0m/3.0s = 13.3 m/s

V(f)^2 = V(0)^2 + 2ad
(13.3)^2 = 0^2 + 2a (40.0)
176.9 = 80a
a= 2.21 m/s^2

Now, a= F(net)/ m
2.21 m/s^2 = F(net)/710
F(net) = 1569.1 N Is that correct???


2. A 225 kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.
F(net) = F(app)- F(res); F(app) = 710 N ; F(f)= F(n)M&; F(n) = mg ??

F(n) = 225(9.8) = 2205
F(f) = (0.20)(2205) = 441

F(net) = 710 N - 441 N = 269 N

A = F(net)/M
A = 269 N/ 225 kg
A= 1.2 m/s^2 ???? I'm not sure I'm getting this...

52. A block lies on a smooth plane tilted at an angle of 22.0 degrees to the horizontal. (a) Determine the acceleration of teh block as it slides down the plane. (b) If the back starts from the rest 9.10m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline? Ignore friction.

Since the verticle component of this is 9.8 m/s^2, 9.8/sin 22.2= 26.2 m/s^2. Thus, the acceleration of the box
is 26.2 m/s^2, but I don't understand how that can be. How can the acceleration down a slope be greater than gravitational acceleration??????? I don't know how to do the next part. I feel like I have a very vague understanding of physics.

THANK YOU!!!!!
 

Answers and Replies

  • #2
Chi Meson
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Originally posted by Astronomer107

1. A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 seconds. The car is uniformly accelerated during the entire time. What net force is exerted on it?

So, acceleration = F(net)/m

V= d/t ; v = 40.0m/3.0s = 13.3 m/s
NO! You can't use the formula for constant velocity if the car is accelerating. Use the formula d=Vot + 1/2 at^2 . Us this to find a directly, then continue as you did with Newton's 2nd law.


2. A 225 kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.
F(net) = F(app)- F(res); F(app) = 710 N ; F(f)= F(n)M&; F(n) = mg ??



F(n) = 225(9.8) = 2205
F(f) = (0.20)(2205) = 441

F(net) = 710 N - 441 N = 269 N

A = F(net)/M
A = 269 N/ 225 kg
A= 1.2 m/s^2 ???? I'm not sure I'm getting this...
That one was exactly right.

52. A block lies on a smooth plane tilted at an angle of 22.0 degrees to the horizontal. (a) Determine the acceleration of teh block as it slides down the plane. (b) If the back starts from the rest 9.10m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline? Ignore friction.

Since the verticle component of this is 9.8 m/s^2, 9.8/sin 22.2= 26.2 m/s^2. Thus, the acceleration of the box
is 26.2 m/s^2, but I don't understand how that can be. How can the acceleration down a slope be greater than gravitational acceleration??????? I don't know how to do the next part. I feel like I have a very vague understanding of physics.

THANK YOU!!!!!
I have to get offline, but this last one suggests that your teacher should be smacked. Hang on for a while...
 
  • #3
Chi Meson
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OK.

In that last one, check your book. THere will be a section where instead of using vertical and horizontal componenets, you instead find the components that are "parallel" and "perpendicular" to the slope of the incline. THis tilted axis is more convenient for problems like this.

So "g" points down, but down is 22 degrees away from the "perpendicular axis." So "g" has a component that is perpendicular and parallel to the incline. THe acceleration down the incline will be the parallel component which is = g sin 22 .
a = 3.7 m/s^2.

THen continue. You actually seem to be getting this more than you think, although I can't believe your teacher plays computer games in class! Are you serious? How does a typical class go for you?
 
  • #4
Originally posted by Chi Meson



So "g" points down, but down is 22 degrees away from the "perpendicular axis." So "g" has a component that is perpendicular and parallel to the incline. THe acceleration down the incline will be the parallel component which is = g sin 22 .
a = 3.7 m/s^2.

[/B]
oh, ok. I think I get it. The diagram shows those lines perpendicular and parallel to the slope, but I think I jsut got confused because gravity had always been the downward component that my teacher never really explained. I have a test tomorrow, so I'm hoping I don't fail. Usually he assigns work in the book (which isn't too clear), then makes us ask questions, but the way he answers them makes us more confused than before. I swear that guy makes up his own equations. The he gives us an assignment and sits at his computer playing solitare. Sometimes we have a pop quiz with 1 or 2 problems that we all never get right I think we really need a teacher who teaches us the material before we are supposed to do it and makes it CLEAR to us. Therefore, I might be here often.... THANK YOU SO MUCH for your help!!!
 
  • #5
enigma
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Originally posted by Astronomer107
The he gives us an assignment and sits at his computer playing solitare.
Ugh. Drop (or mail) an anonymous note off to the principal. Seriously. I hate dead wood.
 
  • #6
Chi Meson
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I'm sorry to hear that. A good physics teacher is hard to get. People who are really good in physics are usually not gifted as teachers. THose who are good at both will usually be teaching at colleges.

And who would in there right mind would pass up an $80,000/year job in favor of a $40,000/year job?
 

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