I'm terrible at M2. Please help!!

For the first one, yes, $\text{Power}=Fv$, but also $\text{Power}_{avg}=\frac{\Delta Work\ Done}{\Delta Time}$



Similarly for the next part you asked about (when Power is constant), $\text{Work}=\text{Power} \times \text{time}$. Does that help?

Finally, you want to find the Work Done by the weight, not just the size of the force. i.e. you need to times 0.05*9.8*850 (the component of weight acting down the plane) by the distance, and then take this away.

If I missed something out, or it didn't make sense, you could ask again. But hopefully that helps .

Thanks for your help.

Your first line. Power = FU? What is F? Force? And U? Velocity?

And AVERAGE power = change in work done divided by change in time?
So in the context of the question (i), what's the CHANGE in work done? From time = 0 to time = 20 seconds?

So work done is 0J at time = 0 seconds, and work done from t = 0 to t = 20 seconds is what? Why is it only the increase in GPE? What about the work done by the driving force of the car up the slope? What about the work done by the car's weight ?

Is it because, since it's travelling at a constant speed, the driving force up the ramp is net 0?

What about the weight though? If we draw a right angled triangle, with the degree of the slope to the horizontal as Y for example, I can see how the distance the car travels is 60/sin(Y) and work done all cancels nicely into the mgh form. But weight is acting DOWN the slope, isn't it? So shouldn't this be negative?
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(iv) work = power*time is meant to help me here?
So I still use 1/2m(v^2-u^2), right?
So that's change in total KE. Which is also work done, isn't it? When you're given power and time you just end up with some figure.

Is it: work done iN THE DIRECTION OF MOTION BY DRIVING FORCE = power * time. Then I include the resistive force too?

What actually is the work done in the equation: work done = PT. Work done by what?

(V) I get your solution perfectly well. But what on Earth is the markscheme donig? What's the 105/20 about? Or their other alternative solution? Both are different to ours, no?

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It's a V. Latex just makes it look weird. i.e. velocity, yes. And F is force. It's what you used.



Average power is just power in this case as it is constant, but that's the correct way to state that equation more generally.

The 'work done' in only the change in GPE in this case because it is travelling at a constant speed i.e. the forces on it are balanced parallel to the slope. The driving force, for example, will do work, but an equal amount of work is done opposing this by the weight acting down the slope. The KE stays the same so the only change in energy is GPE (another way to look at the equation is Power = Energy Transferred/Time). This still probably isn't that good of an explanation, sorry. Hopefully it's a little better though. .


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