Work Energy Principle?

How come the equation for the work energy principle is
Fs= mgh - 0.5 mu^2

Can anyone explain how you get this formula from the other enrgy, work done etc.. equations.
Thanks

Reply 1

phys981

Not entirely sure what you mean, without a question to put it in context, but here goes.

You look at the change in total energy of the vehicle, taking into account its potential and kinetic energy.

If its total energy has increased, the increase must have been provided by the Work Done by a force or engine.

If the total energy has decreased, then this decrease is Work Done AGAINST a resistive force, such as friction.

If that doesn´t help, give me a specific question

Reply 2

phys981

In your equation in post one, I assume Fs is the work done by force F over a distance s.

Let´s suppose the vehicle is travelling uphill. While the vehicle is travelling distance s, it gains a potential energy mgh. Meanwhile it loses ke 1/2 mu^2. If it gains more pe than it loses ke, then overall it has gained (pe - ke).

This energy must be equal to the work done by the driving force of the engine, Fs.

Be careful with the KE.

delta ke is NOT = 1/2 m (delta u)^2

delta ke = 1/2 m ( (u2)^2 - (u1)^2)

It´s a very common mistake to subtract the velocities first then square the difference - that only works of one of the velocities is zero.

I think what I´m trying to say here is your equation in the first post does NOT work if mgh and 1/2 mu^2 are absolute values - they must be the CHANGES that have happened to pe and ke.

Am I making things worse?

Reply 3

mejp44

This is from the June 2004 Edexcel M2 Paper:
In a ski-jump competition, a skier of mass 80 kg moves from rest at a point A on a ski-slope. The skier’s path is an arc AB. The starting point A of the slope is 32.5 m above horizontal ground. The end B of the slope is 8.1 m above the ground. When the skier reaches B, she is travelling at
20 m s1, and moving upwards at an angle  to the horizontal, where tan  = 0.75 , as shown in Fig. 2. The distance along the slope from A to B is 60 m. The resistance to motion while she is on the slope is modelled as a force of constant magnitude R newtons. By using the work-energy principle,

(a) find the value of R.

In the mark scheme they use the formula
Rs= mgh -0.5mu^2
where s=60
m=80,
h= (32.5 -8.1)

Thanks

Reply 4

phys981

OK; over the course of the run, she loses mgh potential energy.

Some of this is converted to ke, the rest is lost as work done against R.

oh ok, cheers

OK, I´ve seen the full question now - the angle of the slop and stuff I think os for the next part.
So should R be 52.2?

Reply 7

-JonBoy-

Hey,

I really need some urgent help with M2!!!

With respect to the work-energy principle, i have seen SO MANY equations relating K.E. P.E. and work done. Below is the one i have found the most frequent:

Increase in K.E + Increase in P.E = W.D by external forces

Can i use this equation for any problem with energy?

is the 'W.D by external forces' negative if the force is in the opposite direction to motion?

If there is a loss in P.E. but a gain in K.E. can i just make the gain in P.E. negative?

Some help would be MUCH APPRECIATED as my exam is this Thursday :frown: