# Work done by F for a particle moving along C

• glid02
In summary, the conversation discusses the calculation of the work done by a radial vector field on a particle moving along a given curve. The work is determined by integrating the dot product of the radial vector field and the derivative of the curve over the entire path. The conversation also mentions the importance of being careful with any missing or incorrect terms in the calculation.
glid02
Here's the question:
If C is the curve given by http://ada.math.uga.edu/webwork2_files/tmp/equations/01/126d71403184689027c86bb27202771.png ,[/URL] http://ada.math.uga.edu/webwork2_files/tmp/equations/56/f3b3a7b9c3e5d804f53afad7a07bc71.png and F is the radial vector field http://ada.math.uga.edu/webwork2_files/tmp/equations/28/71b024e85380a1ae00268ee0af38721.png ,[/URL] compute the work done by F on a particle moving along C.

This is what I've done so far...
F = r because F is only x, y, z
dr/dt = 2cos(t)i+5sin(2t)j+6cos(t)sin^2(t)k

Then F dot dr/dt =
2cos(t)(1+sin(t))+5sin(2t)(1+5sin^2(t))+6cos(t)sin^2(t)(1+2sin^3(t))

Now cos(pi/2) = 0 and sin(pi) = 0 - from 5sin(2t) - so when plugging in pi/2 the answer should be zero for all.
sin(0) = 0 and cos(0) = 1, so when evaluated at 0 the only answer should be 2 (2cos(0)*1) so by my account the answer should be -2.

Apparently this isn't the right answer, if someone could tell me what I missed it'd be awesome.

Thanks a lot.

Last edited by a moderator:
F dot dr is the incremental work along the path... you need to integrate this over the entire path, not just plug in a value!

In this case, W = Int[0, pi/2] F dot dr/dt dt.

Also be careful... you lost a 2 in 2cos(t)(1 + 2sin(t))!

oh wow, that's a great point.

thanks.

## What is work done by a force?

Work done by a force is the product of the magnitude of the force and the displacement of the object in the direction of the force.

## How is work related to force and displacement?

Work is directly proportional to the force applied and the displacement of the object in the direction of the force. This means that the more force applied or the greater the displacement, the more work is done.

## What is the unit of work?

The SI unit of work is joule (J), which is equal to 1 newton-meter (N*m). Other commonly used units include kilojoules (kJ) and calories (cal).

## How is work calculated for a particle moving along a curve?

For a particle moving along a curve, work is calculated by taking the integral of the dot product of the force and the infinitesimal displacement along the curve.

## Can work done by a force be negative?

Yes, work done by a force can be negative if the force is acting in the opposite direction of the displacement. This indicates that the force is doing negative work, or taking away energy from the system.

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