No Work Done during Circular Motion

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Peter G.
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Hi guys, :smile:

Work Done = Force x Distance x cos of Theta - the angle between the Force and the Displacement Vector / direction of motion

In Circular motion I know the force but I am given velocity. In order to prove that no work is done is it correct to say the following:

If we have two velocities, one horizontal and another vertical and find the resultant, the direction of the resultant velocity will be the direction of the displacement vector.

Hence, in circular motion, we only have one velocity vector, we don't need to add. We can then consider it to be the direction of the displacement vector - this would always be at 90 degrees to the force making me understand that no work is done.

Thanks,
Peter G.
 
Ok, so I will try and make my point clearer: Check the attachment.
 
Attachments
  • Slide with the heading "E.x: A ball thrown upwards at an angle" and text about horizontal and vertical velocity, showing a diagram of a blue vertical arrow and a blue horizontal arrow meeting at a right angle, with a red dashed arrow along the hypotenuse as their vector sum.
    Attach.jpg
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