How Do You Calculate the Velocity of Slider A at θ = 90°?

AI Thread Summary
To calculate the velocity of slider A at θ = 90°, the applied horizontal force of 20 N and the gravitational potential energy must be considered. The work done by the force and the change in height contribute to the energy balance equation. Initial calculations incorrectly assumed distances and velocities, leading to confusion. The correct approach involves using the work-energy principle, incorporating both the force and the gravitational potential energy. Ultimately, the calculated velocity of slider A is approximately 3.44 m/s when θ = 90°.
dietwater
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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


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 I am wrong lol can someone explain how i could work this out please
 
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Hi dietwater! :smile:

hmm … your d in your work done is wrong.

And what happened to mgh?
 
tiny-tim said:
Hi dietwater! :smile:

hmm … your d in your work done is wrong.

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 don't know how to get d, I just assumed its 0.4 as its the distance it has to travel?
 
dietwater said:
I don't 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 :wink:
 
tiny-tim said:
No … the point of application of the force is only moving 0.2 :wink:

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
 
Hi dietwater! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

(and your picture didn't show up because you put too many http//s in it :rolleyes:)
dietwater said:
am i right in thinking v(b) is still 2rt2?

Nooo :redface:

imagine it's a cross-roads instead of a T-junction …

which way would B be going as A sails across? :wink:
 
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