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

In summary, when the slider A is released from rest with θ = 0°, it moves with a velocity of 3.44 m/s.
  • #1
dietwater
7
0

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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  • #2
Hi dietwater! :smile:

hmm … your d in your work done is wrong.

And what happened to mgh?
 
  • #3
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?
 
  • #4
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:
 
  • #5
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
 
  • #6
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:​
 

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

1. What is the relationship between work and energy?

Work and energy are closely related concepts in physics. Work is defined as the transfer of energy from one object to another, or the change in energy that occurs when a force is applied over a distance. In other words, work is a measure of the energy that is used to move an object. Energy, on the other hand, is the ability to do work. This means that work and energy are essentially two sides of the same coin - energy is required to do work, and work results in a change in energy.

2. How does velocity affect work and energy?

Velocity, which is the rate of change of an object's position, is directly related to both work and energy. When an object is moving, it possesses kinetic energy, which is the energy of motion. The amount of kinetic energy an object has is dependent on its mass and its velocity - the higher the velocity, the more kinetic energy an object has. In terms of work, velocity also plays a key role. Work is defined as the force applied to an object multiplied by the distance it moves in the direction of that force. Therefore, the faster an object moves, the more work is done on it.

3. How is work and energy related to force?

Work is directly related to force, as it is defined as force multiplied by distance. In order for work to be done on an object, a force must be applied to it. The amount of work done is dependent on the magnitude of the force and the distance over which it is applied. Energy, on the other hand, is related to force through the concept of potential energy. Potential energy is the energy stored in an object due to its position or configuration. For example, a compressed spring has potential energy because it has the potential to do work when released.

4. How is energy conserved in a system with changing velocity?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a system with changing velocity, the total energy will remain constant as long as there are no external forces acting on the system. This means that as the velocity of an object changes, its energy will also change. For example, as an object falls, its potential energy decreases while its kinetic energy increases, but the total energy of the system remains the same.

5. Can an object have work done on it without changing velocity?

Yes, an object can have work done on it without changing velocity. Work is defined as the force applied to an object multiplied by the distance it moves in the direction of that force. If the force is applied in a direction perpendicular to the object's motion, then no work is done on the object. This is because the object does not move in the direction of the force, so there is no change in energy or velocity. However, if the force is applied in the same direction as the object's motion, then work is done and the object's velocity will change.

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