- #1
Fabio010
- 85
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Homework Statement
http://www.freeimagehosting.net/t/7bf5e.jpg
Ok the initial velocity of the car is 5 m/s to the left. Calculate for each situation, the instant time when the velocity of car is :
a)null and 5m/s to the right.
For situation one with just one pulley:
In the block 2 we have F = P-T (=) P = T+m(A)*a (1)
In car, because it is going to the right F = T (=) T = m(B)*a (2)
because he tension that block do in car is equal to the tension that car do in block so
F = P-T (=) F = P - (m(B)*a) (=) P = m(A)*a + m(B)*a (=) P = a(mA+mB)
a = (10*9.8)/(50+10) (=) a = 1.633 m/s^2
So with the law of velocity v = vi + at
v = 0 (null)
vi = 5m/s
a = - 1.633m/s^2 (negative because it is opposite to the car)
t(when v=0) = -5/-1.633 = 3.06s
t(when v = -5) = -10/-1.633 = 6.12s
For the situation with two pulleys
I do not know how to do it, because in the system of the two pulleys we have two tension and i know that they are T/2 each one
but my teacher said that the acceleration of the car is 2* acceleration of the box.
My problem is Why? and even knowing that a(A) = 2*a(B), how can i find the acceleration to the block 3 ?
I appreciate the help.