# Physics Test Help. (teacher recommended internet)

xsrewop27x
Yes AD, that is what I have come to guess and I am attempting to find the solution to the problem by solving it, figuring the data like you did. I am still sitting on a bit of a stump with the problem.

I would like to help with the others but one in particular one caugh my eye. I think i got the 3rd one so correct me if i didnt cause im not in that type of physics, but try the reversal of F=m(a) to find acceleration, f/m=a

Fictional senario but it reales to the problem your doing.

a=Vf - Vi / Time, so 104 m/s - 32 m/s = 72 m/s / 8 sec.
a=9 m/s

Make up a Mass, 1000 Kg

9 m/s x 1000 kg = 9000 N

f/m=a f=9000 N and m=1000 kg

9000 / 1000 = 9 m/s

Back to where we began.
Hope it works, and hope it helps

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Did you solve the problem with the 6 tonne mass on the truck that you posted earlier? I notice you mention five problems in this thread's title, but there are only four here.

Hurkyl
Staff Emeritus
Gold Member
You shouldn't make the same post twice...

xsrewop27x
No I was trying the get that one now. Here is the problem again if you need it.

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?

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

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ShawnD
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:
d = (Vf^2 - Vi^2)/2a
d = (0^2 - 15^2)/2(1.5)
d = -75m
d = 75m

nautica
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

enigma
Staff Emeritus
Gold Member
*sniff*

smells like homework to me.

Doc Al
Mentor
Originally posted by enigma
*sniff*

smells like homework to me.
Yep. And he's posted the same thing in both homework forums.

xsrewop27x
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?

chroot
Staff Emeritus
Gold Member
xsrewop27x,

- Warren

cepheid
Staff Emeritus
Gold Member
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.

Tom Mattson
Staff Emeritus
Gold Member
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: