anonymous
  • anonymous
HEEEELP!! How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart. It first asks what the acceleration will be with no friction is acting on a cart and my formula is: T = m1a m2a = m2g - T T = m2g - m2a m1a+ m2a = m2g a = m2g / (m1 + m2)
anonymous
  • anonymous
and the answer for my current problem stated above is 2.9 m/s^2... and I don't know how to solve it D;

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anonymous
  • anonymous
Someone? Help? :(
anonymous
  • anonymous
a=F/m F=driving force-resistant force what's the driving force in this case?
anonymous
  • anonymous
driving force? It.. It doesn't say.
anonymous
  • anonymous
well i'm just asking you to think about it. it will help you in solving problems in the future. what force is causing the system to move?
anonymous
  • anonymous
ohh.. Tension force.
anonymous
  • anonymous
is this moving to the left?
anonymous
  • anonymous
to the right.
anonymous
  • anonymous
ok, so no it's not the tension force. it's the force from the hanging mass
anonymous
  • anonymous
really? ohh.. okay.
anonymous
  • anonymous
gravity causes the masses to move
anonymous
  • anonymous
ohh, i see it. 'Cause when the hanging mass falls the cart moves.
anonymous
  • anonymous
yup.
anonymous
  • anonymous
so \[driving force=m _{hanging}*g\]
anonymous
  • anonymous
alright.
anonymous
  • anonymous
now the resistant force is already given in the problem
anonymous
  • anonymous
how? And is your equation the same as: a = m2g / (m1 + m2)?
anonymous
  • anonymous
what are you using for m2?
anonymous
  • anonymous
2.0 kg
anonymous
  • anonymous
and you said how the resistant force is already given in the problem-- how?
anonymous
  • anonymous
ok then yes, it will end up as that equation you have but i will break it down a=F/m resistant force = 2N F=2g-2N =19.6N-2N back to acceleration... a=F/mass of the system = (19.6N-2N)/(2kg+4kg)
anonymous
  • anonymous
how did you get 2N as the resistant force? o.O 'cause the answer in the back of the text is 2.9 m/s^2
anonymous
  • anonymous
oops, actually different than yours but i'm confident in what i put up there
anonymous
  • anonymous
do you know a way in which I can use my formula?
anonymous
  • anonymous
my final is 2.9m/s^2
anonymous
  • anonymous
ohhh but do you know how to answer the problem with my formula? ...
anonymous
  • anonymous
tension cancels out for this system which is why i didn't use it in any of my equations
anonymous
  • anonymous
your formula seems a little off but i will double check
anonymous
  • anonymous
ohh.. okay.
anonymous
  • anonymous
break down the system in 2 parts. hanging mass will be labeled 1. \[m _{1}g-T _{1}=m _{1}a\] ^^^derived from F=ma solve for T \[T _{1}=m _{1}(g-a)\] for second part... \[T _{2}-m _{2}g=m _{2}a\] solve for T... \[T _{2}=m _{2}(g+a)\]
anonymous
  • anonymous
ohh..
anonymous
  • anonymous
so what you have to calculate a would work if there was no friction but since there is, the best way to do it is the way i did it
anonymous
  • anonymous
ohh, cos that formula I showed you was for no friction & I thought maybe to switch a few places would be okay.
anonymous
  • anonymous
Well then, thank you so much elica for putting up with me!! Thank you!!
anonymous
  • anonymous
np
anonymous
  • anonymous
btw, why is it: F=2g-2N? Why is it 2g?
anonymous
  • anonymous
after you calculate 2g (2kg*9.8m/s^2) it will equal to 19.6N. the unit N=kg*m/s^2
anonymous
  • anonymous
just remember you can't add or subtract unless they have the same units
anonymous
  • anonymous
ohhh, so it's like m2g?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ohh, okay. Thank you!
anonymous
  • anonymous
np
anonymous
  • anonymous
so F = m2g - frictional resistance? (what's the variable for it?)
anonymous
  • anonymous
in this case, frictional force is denoted as \[F _{k}=\mu mg\] k for kinetic friction and the u-looking sign is the friction coefficient which is usually given in a problem. in this problem, the Fk was already solved (2N)
anonymous
  • anonymous
ohhhh, I'm familiar with that! Oh, thank you!
anonymous
  • anonymous
yup

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