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Not sure about number 3 but i can help with number 4
let a be acceleration , t tension in the string, coefficient of friction = 0.45 equation of motion of block A_ Newtons 2nd law; T - 0.45*2.25g = 2.25 * a equation of motion of block B:- 1.3g - T = 1.3 * a
solve these 2 equations to find T and a
you can find the speed of the bodies after 3 cm by using one of the equations for constant acceleration
you can use g = 9.81N
i got 2.97 for acceleration
is that correct? i got 2 equations from each block then combined those two equations
i got 0.794 for acceleration
and Tension = 11.72N
can i see your work? thanks
i have a question. in block B, i made the equation Tension - Weight = ma
T - 9.933 = 2.25a 12.753 - T = 1.3a adding eliminates T 2.82 = 3.55a a = 0.794 ms-2
yes, but shouldnt be tension positive, and weight is negative? or am i wrong with my analysis?
you have to take account of the friction the frictional force = 0.45 * wieght
no, for block B. there's no friction
the tension is positive for 2.25 block and negative for block B
oh sorry block B force = weight - T
im still confused. haha. wait let me draw
the direction of the tension is upward right, and weight is downward. so i thought its summation of forces would be T-W=ma
or is this analysis wrong?
the resultant force acts downwards weight acts down tensions acts up
so, tension is positive right, if it acts upward?
No it w - t = ma
don't worry too much about positive or negative the point is that the motion is downwards and there for the net force acts downwards
haha, ok. i'll just try it again. but when i get the acceleration, i can just sub it to one of the equations right? to get tension
and then for speed, just use one of the constant acceleration formulas?
yea m use v^2 = u^2 + 2as
u = iniital velocity = 0
sorry i gotta go right now i think you can finish it OK
For number 3 , as the speed is constant I think the force applied will equal the tension in the string. In other words the forces are in equilibrium. what do you think?
wait let me check 3 again
the first thing i did is to make resultant forces = 0 for it is constant speed. is that right?
yes thats what i meant by equilibrium
oh i think i see what you're saying. my solution is not like that. it has frictions
well frictions must play a part
yep but the F is not equal to the tension in my solution
the force must be just enough to overcome the frictions
but the strings on A and B should be equal right? they have the same tension
is the force equal to the sum of the frictional forces on A and B?
that would be 1.8*0.3 + 4.6*0.3?
here's what i did : Block B : Fx = -F + Tension + friction 1 + friction 2 = 0
pretty much yes, but i included tension
yea to be honest i'm not sure of this one.
haha, that's ok i'll try to review with other questions first. haha. thanks though! big help.