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Work energy principle

Is the work energy principle, the total change in energy = total work done on particle

or

Total change in energy = workdone by resistive forces on particle?

How would you apply work energy principle to these situations?
Original post by Zenarthra
Is the work energy principle, the total change in energy = total work done on particle

or

Total change in energy = workdone by resistive forces on particle?

How would you apply work energy principle to these situations?


In these situations it can help to look at things at the begining and at the end.

Try writing down the energy the particle has at each point.

For example, at the point A, what are the values of

1) Kinetic energy 12mv2\dfrac{1}{2}mv^2

2) Gravitational potential mgΔhmg \Delta h

3) Work done overcoming the frictional forces μRd\mu Rd

4) Elastic potential energy, although if this is M2 then it's unlikely to come up, λx2l\dfrac{\lambda x^2}{l}

To answer your initial question, though, what you said first is correct.
Reply 2
Original post by lebron_23
In these situations it can help to look at things at the begining and at the end.

Try writing down the energy the particle has at each point.

For example, at the point A, what are the values of

1) Kinetic energy 12mv2\dfrac{1}{2}mv^2

2) Gravitational potential mgΔhmg \Delta h

3) Work done overcoming the frictional forces μRd\mu Rd

4) Elastic potential energy, although if this is M2 then it's unlikely to come up, λx2l\dfrac{\lambda x^2}{l}

To answer your initial question, though, what you said first is correct.


So if it is total energy change = total work done on particle then for scenario A:

Would it be (mgsintheta + uR)X for total work done on particle

OR

(mgsintheta - uR)X for total work done on particle

OR

(uR)X for total work done on particle
Reply 3
Anyone:P?
Reply 4
My guess for scenario B would be:
Gain in KE = 1/2mv^2 + (Fd-uR)x
Loss in GPE = mgH

Gain in KE - Loss in GPE = workdone by uR?

Any one agree?

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