Jumat, 15 Januari 2010

WORK, ENERGY AND POWER


WORK, ENERGY AND POWER


Imagine a porter carrying a load on his head. Is he doing any work? Yes, one would say! He would be paid for carrying that load from one place to another. But in terms of Physics he is not doing any work! Again, imagine a man pushing a wall? Do you think he is doing any work? Well, his muscles are contracting and expanding. He may even be sweating. But in Physics, we would say he is not doing any work!

If the porter is carrying a very heavy load for a long distance, we would say he has lot of energy. By this, we mean that he has lot of stamina. If a car can run at a very high speed, we say it is very powerful. So, we relate power to speed. However, it means something different in Physics. Let us understand what is meant by work, power and energy in Physics.

Work

Consider the simplest possible case of work done. A force ‘F’ is acting on an object. The object has a displacement ‘S’ in the direction of the force. Then the work done is the product of force and displacement.


What will happen in the case when the applied force is not in the direction of displacement but rather at an angle to it. In such a case we will consider the component of force in the direction of displacement. This component will be effective in doing work as shown.

Component of force in the direction of displacement is F Cos θ.

W = (F Cos θ) S = ∑F. S

Work done is a scalar quantity.

Cases where work is not done

Let us consider some cases where work is not done:
• Work is zero if applied force is zero (W=0 if F=0): If a block is moving on a smooth horizontal surface (frictionless), no work will be done. Note that the block may have large displacement but no work gets done.
• Work is zero if Cos θ is zero or θ = Π/2. this explains why no work is done by the porter in carrying the load. As the porter carries the load by lifting it upwards and the moving forward it is obvious the angle between the force applied by the porter and the displacement is 90o.
• Work done is zero when displacement is zero. This happens when a man pushes a wall. There is no displacement of the wall. Thus, there is no work done. Similarly when a car is moving on a road, there will be a frictional force applied by the road on the. There will be work done by the frictional force on the car. What is the work done on the road by the car? From Newton’s third law we can say that to every action, there is an equal and opposite reaction. Thus the force applied by the road on the car will be equal and opposite to the force applied by the car on the road. Since, there is no displacement of the road, there will be no work done on the road.


Units of Work


Since work done W = F.S, its units are force times length. The SI unit of work is Newton-meter (Nm). Another name for it is Joule (J).

1 Nm = 1 J

The unit of work in cgs system is dyne cm or erg

Note that 1 J = 107 ergs.

The dimensional formula of work is [ML2T-2]

System Unit of work Name of the combined unit
SI Newton meter (Nm) Joule
cgs Dyne centimeter (dyne-cm) Erg

Positive and Negative Work

We have seen the situations when the work done is zero. Work done can also be positive or negative. When 0o <= θ < 90o, work done is positive as Cos θ is positive. Work done by a force is positive if the applied force has a component in the direction of the displacement. When a body is falling down, the force of gravitation is acting in the downward direction. The displacement is also in the downward direction. Thus the work done by the gravitational force on the body is positive. Consider the same body being lifted in the upward direction. In this case, the force of gravity is acting in the downward direction. But, the displacement of the body is in the upward direction. Since the angle between the force and displacement is 180o, the work done by the gravitational force on the body is negative. Note, that in this case the work done by the applied force which is lifting the body up is positive since the angle between the applied force and displacement is positive. Thus work done by the applied force which is lifting up the body is positive since the angle between the applied force and displacement is positive. Thus work done is negative when 90o < θ <= 180o as Cos θ is negative. We can also say that work done by a force is negative if the applied force has a component in a direction opposite to the displacement. Similarly, frictional force is always opposing the relative motion of the body. When a body is dragged along a rough surface, the frictional force will be acting in the direction opposite to the displacement. The angle between the frictional force and the displacement of the body will be 180o. Thus, the work done by the frictional force will be negative. Examples on Work Done

Example 1:
A box is dragged across a floor by a 100N force directed 60o above the horizontal. How much work does the force do in pulling the object 8m?

F = 100N
θ = 60o
S = 8m
W = (F Cos θ) S
=(100 Cos 60o) 8
= 100x1/2x 8 = 400 J

Examples on Work Done

Example 2:
A horizontal force F pulls a 10 kg carton across the floor at constant speed. If the coefficient of sliding friction between the carton and the floor is 0.30, how much work is done by F in moving the carton by 5m?

The carton moves with constant speed. Thus, the carton is in horizontal equilibrium.

F = f = μN = μmg.

Thus F = 0.3 x 10 x 9.8
= 29.4 N

Therefore work done W = FS
=(29.4 Cos 0o)%
=147 J

Variable Force


Force in everyday life is usually not a constant force. So far, when we have discussed work done, we have assumed that the force is constant. What happens if the force is variable? The figure below is a graph of varying force in one dimension.

Let us first understand how de we find work done from a graph. If we are talking about a constant force, the area under the graph (which will be a rectangle) gives us the work done.

Example on Variable Force

Example 3

A force F = 2x + 5 acts on a particle. Find the work done by the force during the displacement of the particle from x =0m to x = 2m. Given that the force is in Newtons.

Work done W = ∫F(x)dx

Thus W = = ∫F(x)dx Cos 0o


= ∫F(x)dx

= ∫(2x + 5)dx

= 2x2/2 + 5x |

= 22 + 5 x 2

= 14 J
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