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?  

Expert Answers
caledon eNotes educator| Certified Educator

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.

adarshanurag | Student

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)

                 = 15(529-225)


                 =4560 J

kavya--kammana | Student

according  to work-energy theorem work done = change in kinetic energy.

`rArr W =DeltaK.E`

`1/2MV^2-` `1/2Mv^2`


`1/2 30(23^2-15^2)`

``=4560 J