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## Homework Statement

http://www.sumoware.com/images/temp/xziltxnrjlckqjqs.png [Broken]

Two blocks ( A and B ) are linked by rope through three identical pulleys C, D, and E.

Pulley C and E are in stationary, while D is pulled down so its constant velocity 1 m/s

If v

_{A}= 4m/s when A goes down 2 m, determine the velocity and acceleration also the direction of block B when A goes 2 m down

Choices :

a) 4 m/s and 1 m/s^2 downward

b) 4 m/s and 2 m/s^2 downward

c) 4 m/s and 2 m/s^2 upward

d) 6 m/s and 4 m/s^2 downward

e) 6 m/s and 4 m/s^2 upward

## Homework Equations

W = ΔKE

V(t)^2 = V

_{0}^2 + 2 a Δs

## The Attempt at a Solution

[/B]

W = ΔKE

ma s = (1/2) m v^2

a (2) = (1/2) (4)^2

a = 4 m/s^2

So, the acceleration of block A is 4 m/s^2 downward

But, what is the block B acceleration and its velocity when block A goes 2m down ?

I need to find the relationship between block A acceleration and block B acceleration.

Their accelerations are same in magnitude, right ? (Since the rope is free)

So, block B acceleration is 4 m/s^2 downward, right ?

But, what about block B velocity ?

Using the equation

V(t)^2 = V

_{0}^2 + 2 a Δs

V(t) = √(2aΔs) = √16 = 4 m/s

But, my answer is not in the choices..

Please help me..

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