# Work and energy with velocity

## 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]

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

Last edited by a moderator:

tiny-tim
Homework Helper
Hi dietwater! And what happened to mgh?

Hi dietwater! And what happened to mgh?

lol il start again with mgh added in! lol

Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

I dont know how to get d, I just assumed its 0.4 as its the distance it has to travel?

tiny-tim
Homework Helper
I dont know how to get d, I just assumed its 0.4 as its the distance it has to travel?

No … the point of application of the force is only moving 0.2 No … the point of application of the force is only moving 0.2 am i right in thinking v(b) is still 2rt2?

If so..

Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

20x0.2 + 2x9.81x0.4 + 0.5x2x(2rt2)^2 = 0.5x2x(v^2)

so v = 3.98

Im not sure where Iv gone wrong

tiny-tim
Homework Helper
Hi dietwater! (have a square-root: √ and try using the X2 tag just above the Reply box )

(and your picture didn't show up because you put too many http//s in it )

am i right in thinking v(b) is still 2rt2?

Nooo which way would B be going as A sails across? 