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Introduction to the concepts of work, energy, momentum and impulse.PowerPoint Presentation

Introduction to the concepts of work, energy, momentum and impulse.

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### Introduction to the concepts of work, energy, momentum and impulse.

F = ma

Ft = mat = mv Ft represents impulse while mv represents momentum. These are all vectorquantities, and as such involve direction.

Fd = mad = m(1/2)v2 where (vf2 -vi2)/2 =ad and vi = 0. Here, Fd represents work and is a scalar quantity as it is proportional to the square of the velocity.

- Note that impulse.work is done when the force and displacement are parallel or anti-parallel.
- Example: Consider the 10 kg mass below which is pulled 5m to the right at a constant velocity.
20N

350 F cos 35

d

Work = W = FcosΘd = (20Ncos35)(5m) = 82 Nm

- Let us now consider a similar situation, which now involves friction. Find the net force acting on the crate, and the work done in moving it a displacement, d, to the right.
N

F

f Θ

Mg

N = Mg – F sinΘ so friction, f, is f = µN = µ(Mg – F sinΘ)

The magnitude of the net horizontal force, which does the work, is

Fnet = F cosΘ - = µ(Mg – F sinΘ) = Ma

W = µ(Mg – F sinΘ) d

Energy is the ability to do work. friction. Find the net force acting on the crate, and the work done in moving it a displacement, d, to the right.Energy of motion is kinetic energy, KE.

W = Fd = mad = m(vf2 -vi2)/2 = ½ mvf2- ½ mvi2

Work is thus equivalent to the change in kinetic energy and is written KE = ½ mv2

Energy due to the position of an object is potential energy, PE or U. Spring potential energy and gravitational potential energy are two common types of potential energies we will consider.

W = Fd = mgd for gravitational PE. So, PE = mgd

Work is thus also equivalent to the change in either potential or kinetic energies, which is known as the Work Energy Theorem

W = ΔKE or W = ΔPE

An application of the theorem: friction. Find the net force acting on the crate, and the work done in moving it a displacement, d, to the right.Consider a 3 kg mass which is sliding down the frictionless incline shown. The angle of the incline is 30 degrees from horizontal , vertical height , h, of the mass is 1m and the length of the hypotenuse of the incline is 2m.Compare the work done by gravity as the mass slides down the incline to the mass done by gravity if the mass simply fell to the ground.

The work done by gravity if the mass falls to the ground is:

W = mgh = (3kg)(9.8 m/s/s)(1m) = 29.4 Kgm2/s2 = 29.4 Nm

The work done by gravity as the mass slides down the incline is:

W = mg(dsinΘ) = (3kg)(9.8 m/s/s)(2m sin 30) = 29.4 Nm

The work done is the same in either case. Thus, the work done is path independent. If we were to raise the mass back to the top; sliding the mass up the ramp and lifting it 1 m straight up would require the same amount of work.

Sign Convention friction. Find the net force acting on the crate, and the work done in moving it a displacement, d, to the right.

- Work can be either positive, negative or zero.
- If the net work done on an object is zero, then the change in kinetic energy is also zero.
- Frictional forces do negative work whenever they act in a direction which opposes motion.
- If force and displacement are parallel, work is positive.
- If force and displacement are anti-parallel, work is negative.
- If force and displacement are perpendicular, zero work is done.

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