Ch 4.2 Help

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Ch 4.2 Help

Post  Admin on Wed Sep 17, 2008 10:32 pm

Question 1: (1 point)

A person's weight is 416.3 N. What is the person's apparent weight when riding in an elevator which accelerates with 0.6748 m/s2 on its way down (see sheet 16,18')? Indicate with a positive (negative) sign whether the result is the same (not the same) if the elevator accelerates with the same acceleration on its way up.

How to solve:

Question 2: (1 point)

Two blocks are connected via a string that runs over a pulley (see sheet 19, replace the 30 degrees by 0 degrees). The 9.236 kg block slides on a frictionless horizontal plane. The other block with a mass of 4.371 kg is suspended from the string. What is the acceleration of the two blocks when they are released (see sheet 23)? Indicate with a negative (positive) sign whether the two blocks have the same (different) acceleration.

How to solve:

Question 3: (1 point)

Two blocks are connected via a string that runs over a pulley (see sheet 19, replace the 30 degrees by 0 degrees). The 4.75kg block slides on a frictionless horizontal plane. The other block with a mass of 5.453kg is suspended from the string. What is the tension of the string when the blocks are released (see sheet 21, eliminate a)?

How to solve:

Question 4: (1 point)

An object moves with constant velocity of 6.672 m/s (both direction and magnitude are constant ). A force of 76.97 N directed west and a second force directed east act on the object. What is the magnitude of the second force (see sheet 6,30)?

How to solve:

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question 1

Post  normansam3 on Thu Sep 18, 2008 12:03 am

for question one take the given weight in newtons and divide by 9.81. this gives you the mass in kg.

now, take the acceleration on earth(9.81) and add it to the elevator. since the elevator is going down, subtract the value.

now take the mass in kg and divide it by modified acceleration.

Mathamaticaly, it looks like this:
416.3 N/9.81m/s*s=42.436kg
9.81m/s*s-.6748m/s*s=9.1352m/s*s

42.436kg*9.1352m/s*s=387.6613472N

the sign should be a negative since going down would require an addition and then the division would be larger, leading to a bigger number.

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Question 1:

Post  Guest on Thu Sep 18, 2008 9:45 am

A person's weight is 416.3 N. What is the person's apparent weight when riding in an elevator which accelerates with 0.6748 m/s2 on its way down (see sheet 16,18')? Indicate with a positive (negative) sign whether the result is the same (not the same) if the elevator accelerates with the same acceleration on its way up.

How to solve:

First find the mass of the person: F = m*a
416.3 = m*9.81(m/s^2)
m=42.4363

Then subtract the elevator's acceleration with gravity since you're going down (in reality you feel lighter when the elevator's going down)
9.81-0.6748 = 9.7425

Then find the new weight with the new acceleration, F=m*a:
F = 42.4363 * 9.7425
F = 413.4365 N
(Answer is negative since the the two values would not be the same since acceleration is different)

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Re: Ch 4.2 Help

Post  normansam3 on Thu Sep 18, 2008 11:03 am

both these solutions for 1 are the same. there is a slight math error in the second explanation hence the deviance in the answers.

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Question 2 and 3

Post  Guest on Thu Sep 18, 2008 8:26 pm

I cant seem to figure out 2 and 3. I have been working on them for a while. If anyone has any idea please reply soon.

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Re: Ch 4.2 Help

Post  Guest01 on Thu Sep 18, 2008 11:05 pm

For Question 2 you want to use the ((m2-m1*sin(x))/(m1+m2)) *9.81

If you are like me and just plug them into a TI-83, don't do it in steps. Find the first half then multiply it by gravity.

m1=9.236kg
m2=4.371kg
g=9.81m/s^2
It tells us that sin(x) is sin(0)
So it would look like something like this
(4.371-9.236*sin(0))/(9.236+4.371)=0.321231719
0.321231719*9.81=3.151283163
rounded to 3.151 as your final numerical answer.

Since the blocks are connected together they function as a single entity, meaning the acceleration of gravity will be the same for both.
So the final answer to enter would be -x.xxx, or in this case -3.151

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Question 3

Post  Guest on Thu Sep 18, 2008 11:22 pm

Any ideas on number 3?

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Re: Ch 4.2 Help

Post  Guest01 on Thu Sep 18, 2008 11:58 pm

It looks like we have to set up a few different things here. Ultimately we are looking for Fnet=ma

It breaks down into this
m2*g=(m2+m1)*a
And solve for "a"

We take that answer and put it into our Fnet=ma equation. Tension is the same thing as Force so T=ma. So set it up and solve.
T=m1*a

Let's try and solve the question now.
m1 = 4.75kg
m2 = 5.453kg
g = 9.81m/s^2
a = __ m/s^2
T (or F) = __ kg*m/s^2 or Newtons (N)

5.453kg*9.81m/s^2=(4.75kg+5.453kg)*a
53.494kg*m/s^2=10.203kg*a
a=5.243m/s^2
Insert into our second equation to find the Tension
T=4.75kg*a
T=4.75kg*5.243m/s^2
T=24.904

Final answer should be 24.904

All numbers rounded to the thousandth decimal place.

Look it over a few times, to get an idea of what's going on

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Re: Ch 4.2 Help

Post  Park on Thu Sep 18, 2008 11:59 pm


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Re: Ch 4.2 Help

Post  Guest01 on Fri Sep 19, 2008 12:21 am

That's where I got the format from.

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Re: Ch 4.2 Help

Post  godlyme on Sat Sep 20, 2008 3:14 am

thanks so much....that help room is worthless

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Re: Ch 4.2 Help

Post  godlyme on Sat Sep 20, 2008 3:18 am

any help with #4

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Re: Ch 4.2 Help

Post  Guest01 on Sat Sep 20, 2008 5:16 am

The magnitude is just the sum of all forces, in summary.

I thought we went over this one, or was that in a separate topic?

You are given two variables, one is velocity and one is a force. Take whatever forces you are given and add them together.

Make sense?

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Tension question

Post  student on Sat Sep 20, 2008 4:44 pm

Two blocks are connected via a string that runs over a pulley . The 4.022 kg block slides on a frictionless horizontal plane. The other block with a mass of 5.96 kg is suspended from the string. What is the tension of the string when the blocks are released

I wrote down
-m_1*g*sintheta+T=m_1*a (sintheta is 0 so ignore/erase -m_1*g*sintheta)
m_2*g-T=m_2*a (add equations together, taking care to get rid of both Ts for the moment)
m_2*g=(m_1+m_2)*a (insert numbers)
(5.96*9.81)/9.982=a
now resubstitute a back into original eqn T=m_1*a
T=4.022*5.857303=23558 which was the correct answer in my pactice set 4_2

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#4

Post  babyElep on Sat Sep 20, 2008 10:00 pm

answer would be the same number as the west direction.

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Re: Ch 4.2 Help

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