- #1

Fabio010

- 85

- 0

## 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:

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.