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## Friday, October 14, 2011

### CBSE PHYSICS Topic-Work and Energy IX

Work
When a force applied on an object and the object moves in the direction of force, we say that the force has done work on the object.

Conditions essential for work to be done are:

1. Some force must act on the object
2. Object move in the direction of force
Hence, the product of the force and the distance moved measures work done.
W = F x S
Where W is the work done, F is the force applied and S is the distance covered by the moving object.
Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account. So work is a scalar quantity

What is the work done when the force on the object is zero?
The work done is zero

What would be the work done when the displacement of the object is zero?
The work done is zero

A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force.  If the force acts on the object all through the displacement, what is the work done in this case?
Work done W = F x S = 5 N × 2 m =10 N m or 10 J.

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force . Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Work done W = F x S = 7 N × 5 m =35 N m or 35 J.

The work done by a force can be either positive or negative.

Work done is negative when the force acts opposite to the direction of displacement.
Work done is positive when the force is in the direction of displacement.

For example when we lift up an object, the force exerted is in the direction of displacement
Hence our work is positive work.  Where as the force of gravity acting on the object try pull towards the earth hence work done by the earth is negative.

A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground.Calculate the work done by him on the luggage.

The work done by porter on the luggage = F S = mg x s = 15 kg × 10 m s-2 × 1.5 m
= 225 kg m s-2 m    = 225 N m = 225 J

If a force is applied at an angle θ to the displacement of the particle, the resulting work is defined by: FS cos q

Let a constant force F acting on a body produce a displacement S as shown in the figure. Let q be the angle between the direction of the force and displacement.

Displacement in the direction of the force = Component of S along AX = AC
We know that Cos q    = adjacent side/ hypotenuse
Cos q = AC/S
AC= S Cosq

Displacement in the direction of the force = S cos q
Work done = Force x displacement in the direction of force
W = FS cos q
The displacement S is in the direction of the force F, and
If q = 0, cos 0o =1
Then, W = FS x 1
W = FS
If q  = 90o, cos 90o = 0
Therefore, W = FS x 0 = 0 i.e., no work is done by the force on the body.

SI Unit of Work
W = F x S    SI unit of F is N and that of S is m [N = Newton]
SI unit of Work =Nm
1N m is defined as 1 joule.   i.e., 1 joule = 1 N m
So SI unit of Work is Joule

1 N m is referred as joule after the British Scientist James Prescott Joule .The letter 'J' denotes Joule.   Higher units of work are kilo joule and mega joule.
1 kilo joule = 1000 J    or 102 J
1 mega Joule = 1000000 J = 106 J

One joule is the work done when a force of one Newton moves an object through a distance of one meter in the direction of force.

Question  1. When do we say that work is done?

Ans :   Work is done whenever the given conditions are satisfied:
1. Some force must act on the object
2. Object move in the direction of force

Question  2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Ans: When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:
Work done = Force × Displacement
W = F × s

Question  3. Define 1 J of work.
1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force

Question 4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Ans: Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W = F × s = 140 × 15 = 2100 J  Hence, 2100 J of work is done in ploughing the length of the field.

Question 5:A 10 kg object experiences a horizontal force which causes it to accelerate at5 m/s2 , moving it a distance of 20 m, horizontally. How much work is done by the force?
Solution :  F = ma = (10)(5) = 50 N.
It acts over a distance of 20 m, in the same direction as the displacement of the object,
So, The total work done by the force
W =  F x S = (50)(20) = 1000 Joules.
Question 6. A ball is connected to a rope and swung around in uniform circular motion. The tension in the rope is measured at 10 N and the radius of the circle is 1 m. How much work is done in one revolution around the circle? fig. 02
Solution: In uniform circular motion, centripetal force is always directed radically, or toward the center of the circle. Also, the displacement at any given time is always tangential, or directed tangent to the circle
So,  the force and the displacement will be perpendicular at all times. and  q = 900
W = F s cos θ , (Cos  900  = 0 )
No work is done on the ball.

Question 7. A  crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. The tension in the rope is 50 N. How much work is done in moving the crate 10 meters?
Solution : In this problem a force is exerted which is not parallel to the displacement of the crate. Thus we use the equation W = F s cosθ .= (50)(10)(cos 30) = 433 J

Question 8. A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight?
Solution: The crate, and thus the point of application of the force, does not move. Thus, though a force is applied, no work is done on the system.

Question 9. A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is  .25.  How much work is done by the 50 N force in moving the block a distance of 10 meters? What is the total work done on the block over the same distance?
Solution :  Finding the work done by the 50 N force is quite simple. Since it is applied parallel to the incline,
The work done is simply W = Fx = (50)(10) = 500 J.
Finding the total work done on the block is more complex.
The first step is to find the net force acting upon the block. To do so we draw a free body diagram:fig. 03
Because of its weight, mg, the block experiences a force down the incline of magnitude mg sin 30 = (5)(9.8)(.5) = 24.5 N .

In addition, a frictional force is felt opposing the motion, and thus down the incline. Its magnitude is given by F k =  μ F N = (.25)(mg cos 30) = 10.6 N .

In addition, the normal force and the component of the gravitational force that is perpendicular to the incline cancel exactly.

Thus the net force acting on the block is: 50 N -24.5 N -10.6 N = 14.9 N , directed up the incline.

It is this net force that exerts a ìnet workî on the block.

Thus the work done on the block is W = Fs  = (14.9)(10) = 149 J.

Energy
Any thing that is able to do work possesses energy. Energy is the capacity to do work. Energy is measured by the amount of work that a body can do. Therefore, SI unit of energy is also joule.

How does an object with energy do work?

An object that possesses energy can exert a force on another object to do work.

When this happens, energy is transferred from one object to the other. The second object may move as it receives energy and therefore do some work. Thus, the first object had a capacity to do work. This implies that any object that possesses energy can do work.

Kinetic energy and potential energy are the two types of mechanical energy.

Kinetic energy is defined as the energy possessed by an object by virtue of its motion.

Kinetic energy is represented by the letter K. All moving objects possess kinetic energy.

Expression for kinetic energy
Let the force 'F' does work when it moves a body through a distance 'S' and this work done is stored in the body as its kinetic energy.

By definition, W = F x S  --------------------------(1)
F = ma                     [Newton's second law of motion]

So, W = ma s    --------------------(2 )

Also,       v2 - u2 = 2as         [Newton's third law of motion]

v2 - 0 = 2aS         [Initial velocity u = 0 as the body is initially at rest]
v2 = 2aS
a = V2 /2s
By substituting the value of 'a' in equation (2) we get,

W = m v2 s / 2s = m v2 / 2 ------------------ (3)

Since work done is stored in the body as its kinetic energy , We can write equation as

Kinetic energy = 1/2 m v2

we can conclude that the kinetic energy of a body is directly proportional to
(1) its mass                   and                         (2) the square of its velocity.

Activity
Take a heavy ball and drop the ball on the sand bed from height of about 25 cm , 50 cm, 1m and 1.5 m.. The ball creates a depression. Ensure that all the depressions are distinctly visible.

Which one of them is deepest? Which one is shallowest? Why?
Ans: a ball drop from 1.5 m is deepest and a ball drop from 25 cm is shallowest because a ball drop from 1.5 m posses more kinetic energy.

Momentum:  Momentum (p) of the body is defined as the product of its mass and the velocity.
Let a body of mass 'm' moving with a velocity 'v'.
Then, momentum of the body is got by   p = m v  Þ V =p/m
But,  Kinetic energy = 1/2 m v2
Substituting the value of v in equation
Kinetic energy = 1/2 m (p/m) 2 =  p2 / 2m

Law of Conservation of Energy

Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another

Question 1: What is the kinetic energy of an object?

Ans: kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object increases with its speed. Example:  A body uses kinetic energy to do work. Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run wind mills, etc.

Question 2 : How much energy is possessed by a moving body by virtue of its motion?

Ans: The kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it acquire that velocity.

Questions: An object of mass 15 kg is moving with a uniform velocity of 4 m/s. What is the kinetic energy possessed by the object?

Solution: Kinetic energy = 1/2 m v2  =1/2 x 15  kg × (4 m/ s)2 = = 120 J
The kinetic energy of the object is 120 J.

Question:  What is the work to be done  to increase the velocity of a car from 30 km/ h to 60 km/ h if the mass of the car is 1500 kg?
u = 30 km/ h = 8.33m/s   v = 60 km/ h = 16.67 m/s
The initial kinetic energy of the car= 1/2 mu2 = 1/2 x 1500 kg × (8.33 m/ s)2 = = 52041.68 J.
The final kinetic energy of the car = 1/2 mv2 = 1/2 x 1500 kg × (16.67 m/ s)2  = 208416.68 J.
Thus, the work done = Change in kinetic energy = 208416.68 J. 52041.68 J = 156375 J.