KINETIC ENERGY THE ENERGY POSSESSED BY A BODY BY

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Kinetic Energy


Kinetic energy

The energy possessed by a body by virtue of its motion called the kinetic energy of the body. A rocket traveling to the Moon has kinetic energy as does a snail crawling along a wall. It is the kinetic energy of objects that makes them difficult to stop and the kinetic energy of the air in a hurricane that causes severe damage to the countryside.


The kinetic energy of an object depends on two things:

(a) the mass of the object (m)

(b) its speed (v)


The formula for kinetic energy of an object of mass m travelling at velocity v is:

Kinetic energy (k.e) = ½ mv2





Proof of the formula for the kinetic energy of an object

Suppose that a body of mass m initially at rest is accelerated by constant force F so that it reaches a velocity v after travelling a distance of s metres. (Figure 1).

TKINETIC ENERGY THE ENERGY POSSESSED BY A BODY BY
he kinetic energy gained by the body is equal to Fs, and this must therefore be the work done in giving it a velocity v.


However we know that F = ma and v2 = u2 + 2as where u and v are the initial and final velocities of the object. In this case u = 0 and so we have:


v2= 2as and so ½v2 = as


Therefore: kinetic energy = Fs = mas = ½ mv2


Kinetic energy = ½ mv2




The kinetic energy of an object is measured in Joules (J), kilojoules (kJ) or Megajoules (MJ).


Estimate of the kinetic energies of various objects:


speeding car 650 kJ

sprinter 5 kJ

oil tanker 5000 MJ

air molecule 2.5x10-21 J






Kinetic energy changes

It is important to understand the correct way to calculate changes in the kinetic energy of an object.


For example suppose we want to find the increase in the kinetiec energy of an 8kg ball when its velocity is increased from 3 ms-1 to 4 ms-1. The correct way is as follows:

KINETIC ENERGY THE ENERGY POSSESSED BY A BODY BY

Kinetic energy increase (ke) = ½ x8x[42 – 32] = 4x[16-9] = 4x7= 28 J


and NOT


K

Example problem

A lorry of mass 6000 kg travels along a level road a 30 ms-1. The brakes are then applied and the lorry stops in 70 m. Calculate:

(a) the kinetic energy of the lorry before braking

(b) the braking force


(a) kinetic energy = ½ mv2 = ½ x 6000 x 302 = 2 700 000 J = 2.7 MJ

(b) braking force = Energy changed/ Braking distance = 2.7x106/70 = 38.6 kN



inetic energy increase = ½ x8x[4-3]
2 = 4 J

2



A COMPARISON OF SELECTED ISOKINETIC TRUNK STRENGTH PARAMETERS OF
ADVANCED PHARMACOKINETICS PHA 833 3 CR ADVPK2PHD COURSE DESCRIPTION
ANSWERS FOR THE PRACTICAL EXERCISE KINETIC MODEL FITTING (PARENT


Tags: energy the, inetic energy, energy, possessed, kinetic