Physics Test Help. (teacher recommended internet)

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Homework Help Overview

The discussion revolves around a set of physics problems presented in a take-home test for 11th grade students. The problems involve concepts related to inclined planes, tension in strings, and forces acting on objects. Participants express confusion and seek clarification on specific questions while adhering to collaboration guidelines set by the teacher.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the forces acting on objects on inclined planes and the relationship between tension in strings and weight. Some express confusion about specific problems and seek diagrams for better understanding. Others share their reasoning and calculations related to the problems.

Discussion Status

Some participants have provided insights and partial solutions to the problems, while others are still grappling with understanding the concepts. There is a mix of attempts to clarify the problems and share reasoning, but no consensus has been reached on the solutions. Guidance has been offered regarding the approach to certain questions, particularly in understanding the mechanics involved.

Contextual Notes

Participants note the constraints of the homework assignment, including a tight deadline and the requirement to collaborate without direct solutions. There is mention of diagrams that may aid in understanding, and some participants express the need for further assistance with additional problems beyond those initially posted.

  • #31
Well, the force on the object cannot be greater 9000N when it decelerates. So the maximum deceleration is when F = 9000N.

F = ma

9000 = 6000a

a = 9000/6000 = 1.5ms-2

v2 = u2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

The acceleration is negative, as it is decelerating and the final velocity is zero, so

u2 = 2as

And

s = u2/2a

So

s = u2/3 = 225/3 = 75 m
 
Last edited:
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  • #32
Originally posted by xsrewop27x
An object of mass 6000 kg rests on the flatbed of a truck. It is held in place by metal brackets that can exert a maximum horizontal force of 9000 N. When the truck is traveling 15 m/s, what is the mimimum stopping distance if the load is not to slide foward into the cab?

maximum acceleration the brackets can hold:
F = ma
a = F/m
a = 9000/6000
a = 1.5m/s^2

acceleration compared distance:
Vf^2 = Vi^2 + 2ad
2ad = Vf^2 - Vi^2
d = (Vf^2 - Vi^2)/2a
d = (0^2 - 15^2)/2(1.5)
d = -75m
d = 75m
 
  • #33
Maybe, your teacher is wanting you to use your own mind to learn the material, which is most cases is a much more effective way of teaching. If you are in high school, then you are probably used to being food feed the information, which will do nothing for your future.

Wether this teacher is trying to make you use your own mind or he simply can not teach the material is irrelevant. Either way, you will be better off, by taking the time to learn and understand the material for yourself.

Repost each question one at a time in the homework section, but before posting it - try to work the problem and post your solution. We can help you figure it out from there.

Good luck
Nautica
 
  • #34
*sniff*

smells like homework to me.
 
  • #35
Originally posted by enigma
*sniff*

smells like homework to me.
Yep. And he's posted the same thing in both homework forums.
 
  • #36
No these are my answers .. are they right?
1)75m

2)2kg

3)96.5N

4)2 m/s2

5)6.18N

Do you all agree that I got the right answers?
 
  • #37
xsrewop27x,

Please don't crosspost, and please don't ask us to do your homework for you. Since you already have an active thread there in Homework Help, I'm closing this one.

- Warren
 
  • #38
Let's look at # 1:

At first (using Newton's 2nd Law),

a1 = F/m1 = 3.0m/s^2

F = 3m1

And then,

a2 = F/m2 = 2.0m/s^2
F = 2m2

but we know that m2 = m1 + 1, so 2m2 = 2(m1 + 1) = 2m1 + 2

F = 3m1 = 2m1 + 2 (solve for m1)

m1 = 2kg

Pretty simple, actually.
 
  • #39
This thread is going to look a little screwy because It is actually 4 identical threads merged (by me) into a single thread.

xsrewop27x:

First, it is considered spamming to post identical threads more than once. Second, while we are happy to help you with your homework, we will not do it for you. That said, we require that you show how you started each problem, and we will help you with where you get stuck. Give each one a try--they are not very difficult.

Others:

Please stop posting complete solutions.

Thanks,
 

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