A 30 kilogram mass is moving at 15 meters per second on a horizontal, frictionless surface. What is the total work that must be done on the mass to increase its speed to 23 meters per second?
Work is normally expressed as Fd, or Force acting over a distance. Force is defined as mass accelerating, or ma. Putting the two together we have mad. However we can't really work with this because we only have a mass, M. We don't know acceleration because we haven't been given a distance or a time over which the force accelerates the mass. We need to work around this. Conveniently, Work and Kinetic Energy are both expressed in Joules. We will use this as a starting point.
We need an equation which makes use of the variables that we DO have: mass and velocity. Kinetic energy is the term for energy derived from physical motion. .5mv^2 is the formula for KE.
The mass has an initial kinetic energy of .5(30)15^2, or 3375J.
It has a final kinetic energy of .5(30)23^2, or 7935J.
The total amount of kinetic energy added is 7935 - 3375 = 4560 joules.
So 4560 joules of work need to be done on the mass in order to accelerate it from 15m/s to 23m/s. Note that it doesn't matter how much time this takes or over what distance it occurs: if Work = mad, and we only know that the value of work is 4560 and the value of m is 30, then a and d can be any value inversely proportional to each other whose product equals 152. For example: let a = 1m/s and d = 152m
Vf^2 = Vi^2 + 2ad
23^2 = 15^2 + 2(1)(152)
529 = 225 + 304
529 = 529
We could completely reverse this and let a = 152m/s^2 and d = 1m, and the result would be the same.
According to work energy theorem, work done on a moving body is equal to the gain in kinetic energy of the body.
K.E = 0.5*m*v*v
New K.E = Work done + old K.E
Work done =new K.E - old K.E
= 0.5*m*V*V - 0.5*m*v*v
=0.5*m(V*V - v*v)
= 0.5*30(23*23 - 15*15)
according to work-energy theorem work done = change in kinetic energy.
`rArr W =DeltaK.E`