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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
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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)
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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Other answers:

Someone? Help? :(
a=F/m F=driving force-resistant force what's the driving force in this case?
driving force? It.. It doesn't say.
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?
ohh.. Tension force.
is this moving to the left?
to the right.
ok, so no it's not the tension force. it's the force from the hanging mass
really? ohh.. okay.
gravity causes the masses to move
ohh, i see it. 'Cause when the hanging mass falls the cart moves.
yup.
so \[driving force=m _{hanging}*g\]
alright.
now the resistant force is already given in the problem
how? And is your equation the same as: a = m2g / (m1 + m2)?
what are you using for m2?
2.0 kg
and you said how the resistant force is already given in the problem-- how?
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)
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
oops, actually different than yours but i'm confident in what i put up there
do you know a way in which I can use my formula?
my final is 2.9m/s^2
ohhh but do you know how to answer the problem with my formula? ...
tension cancels out for this system which is why i didn't use it in any of my equations
your formula seems a little off but i will double check
ohh.. okay.
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... \[T _{2}=m _{2}(g+a)\]
ohh..
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
ohh, cos that formula I showed you was for no friction & I thought maybe to switch a few places would be okay.
Well then, thank you so much elica for putting up with me!! Thank you!!
np
btw, why is it: F=2g-2N? Why is it 2g?
after you calculate 2g (2kg*9.8m/s^2) it will equal to 19.6N. the unit N=kg*m/s^2
just remember you can't add or subtract unless they have the same units
ohhh, so it's like m2g?
yes
ohh, okay. Thank you!
np
so F = m2g - frictional resistance? (what's the variable for it?)
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)
ohhhh, I'm familiar with that! Oh, thank you!
yup

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