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The Power

The rate of doing work is called power, i.e.


The faster a given amount of work is done, the greater is the power of the agent that does the work. In the SI system, the unit of power is the watt (symbol W). Power is said to be 1 watt when 1 joule of work is done in 1 second, i.e.
1 W = 1 J s-1

Since the watt is a small unit for the measurement of power, larger units, namely kilowatt (KW) and megawatt (MW) are often used.
1 kW = 1000 W = 103 w
1 MW = 1,000,000 W = 106 W
1 MW = 1,000,000 W = 106 W

The power of an agent can also expressed in terms of the force applied and the velocity of the object on which the force is applied. Now, power P is given by
 
P = F· v

Power is a scalar quantity as it is the ratio of two scalars W and t, or a scalar product of two vectors F and v.

Example
A toy rocket of mass 0.1 kg having a small quantity of fuel of mass 0.02kg is at rest on a horizontal smooth track. When ignited, its fuel burns out in 3 s and it acquires a speed of 20 ms-1 after all the fuel is burnt. If the small mass variation of the rocket during the burning of the fuel is ignored, find (i) the thrust on the rocket, and (ii) the energy content per unit mass of the fuel.

Solution  
Mass of the rocket (m) = 0.1kg, 
Initial velocity (u) = 0, 
Final velocity (υ) = 20ms-1 and time taken (t) = 3 s.
  1. The acceleration of the rocket is

  2. The energy content of the fuel = work done by the fuel to give a speed υ = 20 m s-1 to the rocket in time t = 3s. If S is he distance covered by the rocket during this time, then


    Hence energy content per unit mass of fuel =  

Example
A block of mass 2.0 kg initially at rest on a horizontal floor moves under the action of an applied horizontal force of 10N. If the coefficient of kinetic friction between the block and the floor is 0.2, find the
(i) work done by the applied force is 5 s, 
(ii) work done by friction in 5 s,
(iii) work done by the net force on the block in 5 s, and
(iv) the change in the kinetic energy of the block 5 s. What conclusions would you draw from your results?

Solution 
m = 2.0 kg, u = 0,F = 10 N and μ = 0.2
Now
Force of friction f = μ mg = 0.2 2.0 9.8 = 3.92 N
Since the friction opposes motion, the net force acting on the body when it is moving is
Fn = F-f = 10-3.92 = 6.08 N
The acceleration produced in the block is

The distance moved by the block in t = 5 s is

  1. Work done by the applied force in 5 s is
    W = applied force distance moved in 5 s
    W = F S
    W = 10 38 = 380 J
  2. Work done by the force of friction in 5 s is
    W1 = f s = 3.92 38 = 148.96 J
  3. Work done by the net force in 5 s is
    W2 = Fn S = 6.08 38 = 231.04 J
  4. velocity (v) acquired by the block in 5 s is given by
    υ = u + at = 0 + 3.04 5 = 15.2 m s-1
    ∴ Kinetic energy of the block at t = 5 s

    Initial kinetic energy of the block is
    ∴ Change in kinetic energy = 231.04 J
Conclusions We find that W1 + W2 = 148.96 + 231.04 = 380 J which is W, the work done by the applied force. This means that a part = W1 = 148.96 J of the work W is used up in overcoming the friction between the block and the floor and remaining part W2 = 231.04 J is the useful work done by the force. The block moves at the expense of this work.

Furthermore, the useful work done = 231.04 J equals the change in the kinetic energy of the block. This is the work-energy principle which states that the work done by the net force acting on a body is equal to the change in its kinetic energy.




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