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## Homework Statement

Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its

respective guide, with y being in the vertical direction (see Figure 3). A 20 N horizontal force

is applied to the midpoint of the connecting link of negligible mass, and the assembly is

released from rest with θ = 0°. Determine the velocity vA with which slider A strikes the

horizontal guide when θ = 90°.

[vA = 3.44 m/s]

http://http://img.photobucket.com/albums/v242/021003/tutorial12.jpg [Broken]

## Homework Equations

1/2 mv^2

F = ma

Wp = mgh

SUVAT

## The Attempt at a Solution

When at 0 degrees

W=0J

At 90

F=20N

W = 20xd = 8J

Work from cart A = 0.5mv^2

Therefore 16 = mv^2

v = 2 rt2

Or...

do i need to add the energy from 20n force and from cart b...

0.5mv^2 (b) + 8J = 0.5mv^2 (A)

with F = ma, 20/10 a = 10 therefore v (b) = 2 rt 2

sub this into above eq.

8 + 8 = 0.5mv^2

v = 4

Help!

Iv been goin round in circles, clearly im wrong lol can someone explain how i could work this out please

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