Work, Energy And Power SS2 Further Mathematics Lesson Note

WORK

If the point of application of a force is displaced, the force is said to do work. For a constant force F, whose point of application is given a displacement d, work done is the product of /F/ and d. 

The work done by a constant force is defined as the product of the force and the distance moved by its point of application along the line of application of the force.

W= /f/d

If the magnitude of the displacement is d then 

W= /F/ × d

POWER 

Power is the rate at which work is being done. For example, if work of 90J is done in 15 seconds, the power is 6J/sec. The unit of power is the Watt (W)

Example:

On the level, a car develops a power of 60KW. If the resistance to motion is 9ooN, what is the maximum speed of the car?

Working at the same power and with the same resistance operating, what would be the maximum speed possible up an inclined plane whose slope is sin-1(1 )

                      50, if the mass of the car is 800kg?

What is the acceleration at the time when the car is moving up the inclined plane at 40 ms-1? (Take g= 10ms-2)

Solution 

Let P be the power developed by the car.

Let W be the work done.

Let F be the tractive force of the car

P = d  (w)

      dt

= d (f × s)

   dt

ENERGY

1. Kinetic Energy

The work done in bringing a particle of mass m from rest to a velocity v is called the kinetic energy of the particle

If we donate Ek as the kinetic particle of mass m reaching a velocity v from rest, then

Ek =½mv²

If W is the work done in bringing the particle of mass m to a velocity v from an initial velocity u, and if s is the distance travelled in the process, then

W= F × S

=m × a × s where a is the acceleration

But

s = (u + v)t

          2

a= (v – u)t

         2

:. W = m × a × s

= m × (v-u) × (v+u)t

              t           2

=½mv² – ½mu²

If we denote the last expression b

∆Ek = ½mv² –  ½mu²

∆Ek is the change in kinetic energy in bringing a particle of mass m from an initial velocity u to a final velocity v

Potential energy

The potential energy of a particle of mass m is the energy of the particle, by its position relative to a reference level. It is the work done in bringing a particle from a reference level to a height h

If we donate the potential energy by Ek then 

Ep = f × s

Since F = mg, and S = /

Ep= mgh

Law of Conservation of Energy 

This is a generalization of the experience from nature and it states that energy cannot be created nor destroyed. It can only be changed from one form to another.

When a particle falls from a height, it loses potential energy. This loss in potential energy is compensated for, by the gain in kinetic energy.

Example 

A particle starting from rest falls freely from a height H above the ground. If g is the acceleration due to gravity, show from energy consideration that the velocity v with which the particle strikes the ground is given by the expression 

v = ✓2gH

Solution 

Let the particle have a mass mkg.

Loss of potential energy = mgH.

Gain in kinetic energy = ½mv²

ASSIGNMENT 

1) A body of mass 20kg moves a distance of 8m in the direction of the line of action of force F =5N  on a smooth table find the work done

2) A particle of mass 3kg is projected vertically upward with an initial velocity of 5 m/s from the ground, calculate the potential energy at the greatest height

3) A sphere of mass 12 kg and another sphere of 8kg move toward each other with velocities 5 m/s and 3 m/s respectively find the speed of the sphere after collision

4) Calculate the loss in kinetic energy caused by the collision of the two bodies above in (3)