Summer Term

Spring term archive

11/05/09

Going through papers - bring questions please!

Amir! Sort out bibliography!


07/05/09

Go through/do of the mechs test. Then on to actual exam papers...

Dayan and Amir bibliography?


05/05/09

Mechanics test will have been sat.


30/04/09

This time you did do the relevant work, energy and power questions. We then looked at stepwise calculations of motion where the motion is broken up into small time steps over each of which motion equations can be applied. This kind of iterative calculation as a model of reality can lead to errors which are aggravated when the time interval chosen is too large.

HW Revise mechanics for a test on Tuesday (Monday timetable.)


27/04/09

No RM - on a course. You guys should have done Work, energy power exam Qs but they weren't given out. Poo.


23/04/09

We ploughed on with the last few bits of theory to do. I realised that I had missed out the section on the "step by step" calculation of the path of an object. We'll have a look at this next time.

We went through some work, energy and power ideas. Firstly, we recapped Newton's 3 laws and looked at the idea of impulse. A certain change in momentum can be caused by a small net force acting over a long time or a large force acting over a short period of time - they have the same impulse.

We then looked at how energy is used in mechanics problems. Firstly we recapped that work done by a force is dependent on the distance moved in the direction of the force.

Work done = Force times displacement (in the same direction as the force)

You may be called upon to use trig. to resolve vectors in order to work out the component of a displacement which is in the same direction as the line of action of the force.

So forces do work - in mechanics they often cause accelerations - e.g. give mass some kinetic energy.

Work done = Fd

d is displacement - x.

x = ut + 1/2at2

u for a body starting at rest is zero.

W = 1/2Fat2

a = (v-u)/t and u is still zero.

W = 1/2Fvt

F = ma = m(v-u)/t

u = 0

W = 1/2mv2

We also looked at the transfer of GPE (GPE = mgh) into KE, both in direct falling situations and down slopes.

Any discrepancy between the original GPE of an object and its subsequent KE after falling can be used to make a calculation of the average frictional force acting on the object during its fall. (using W=Fd)

For an abject travelling at a constant velocity against a constant frictional force, it can be shown that the power required is: P = Fv

F = frictional force

v = velocity

HW Yes - from the book I feel. P 215 all Qs please.


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