Physics Test Help. (teacher recommended internet)

Click For Summary
SUMMARY

This forum discussion revolves around a high school physics test that includes challenging problems related to inclined planes, tension in strings, and friction. The user seeks assistance with specific questions, including calculating acceleration on a frictionless inclined plane at 30 degrees, determining tension in strings supporting a lantern at 35 degrees, and finding the maximum angle for a sled on a hill with a coefficient of friction of 0.10. The solutions provided utilize fundamental physics principles such as Newton's Second Law and trigonometric relationships.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Familiarity with concepts of friction and equilibrium
  • Ability to interpret and analyze free-body diagrams
NEXT STEPS
  • Study the mechanics of inclined planes in physics
  • Learn how to derive tension in systems with multiple strings
  • Explore the concept of limiting equilibrium in friction problems
  • Practice solving problems involving forces on inclined surfaces
USEFUL FOR

High school students preparing for physics exams, educators looking for teaching resources, and anyone seeking to improve their understanding of mechanics and problem-solving in physics.

  • #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:
Physics news on Phys.org
  • #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,
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
7K
Replies
10
Views
4K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
30
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K