How much work is done when a force of 20 N displaces a body by 10 m in the direction of the applied force?

In our day to day life, the word work means any kind of mental and physical activity. Work is said to be done when a force displaces a body through certain distance in the direction of force.For example, we say that we are doing work while,

•  reading a book,
• cooking the food,
• A boy pulling a toy car with a string. The change in position of the toy car shows that some work has been done.
• walking on a level road with a box on our head,
• pushing a wall of a house but fails to do so.
•  A bullock pulling a cart. The cart moves. The bullock pulls the cart with a force  which moves the cart in the direction of force and hence the work is said to be  done.

In all these cases, either mental or physical activity is involved.

But is physics, the term work has entirely a different meaning. In physics work is done if a force applied on a body displaced the body in its own direction. In other words, the condition which must be satisfied for the work done are : (i) a force must act on the body and (ii) the body must be displaced from one position to another position. Thus, no work is done in (i) , (ii) ,(iv) ,(v) cases mentioned above. But in the case (ii) and (vi) when a force displaces a body through certain distance in the direction of force work is said to be done.

DEFINITION:

Work is said to be done when a force displaces a body through certain distance in the direction of force.

For example :

• A batsmen uses a bat to hit a ball when he wants to hit  the ball six.
• A bullock pulling a cart. The cart moves. The bullock pulls the cart with a force which moves the cart in the direction of force and hence the work is said to be done.
• A man pushing a car.

Thus, the work done by a force on a body depends on two factors :

• Magnitude of the force.
• Distance through which the body moves in the direction of force.

In other word the work done by a given force on a body depends only on the force, the displacement, and the angle between them.  It does not depend on the velocity or the acceleration of the body, or on the presence of other forces. 

MEASUREMENT OF WORK:

Let a constant force, F acts on an object and displaces the object through a distance‘s’ in the direction of the force applied. Then, the work done W is given by the product of force exerted on the body and the distance moved by the body in the direction of force. i.e.

Work = Force × Displacement

W = F ×s

Work done has only magnitude and no direction. Therefore, it is a scalar quantity.

If F = 1 N and s = 1 m, then work done by the force will be 1 Nm.

The S. I. unit of work is Joule or Nm.

1 Joule is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of force.

The condition for a force to do work is that it should produce motion in an object.

If however the distance moved is zero, then the work done on the object is always zero.

 Units of Work done :

Work done, W = Fd

In C.G.S. system the unit of work done is dyne x cm = erg.

Definition of 1 etg :

If F = 1 dyne and d = 1 cm.

then W = 1 × 1 = 1 erg.

If one dyne force is applied on a body and displacement in the body in 1 cm in the direction of force, then work done will be one erg.

S.I. unit of work done is Newton × metre = joule.

Definition of 1 joule :

if F = 1N and d = 1m.

then, W = 1 × 1 = 1 joule (J)

If a force of 1 Newton is applied on a body and displacement in the body is 1m in the direction of force then work done will be 1 joule.

Relation between joule and erg :

1 joule = 107 erg

Work is measured by the product of force and the displacement in the direction of force. Work is a scalar quantity.

Work = Force × displacement in the direction of force

W = F(d cos

or work done = displacement × force in the direction of displacement.

W = d(F cos

Things to remember :

Gravitational unit of work:

Work is said to have gravitational unit of work if unit gravitational force displaces the body through unit distance in the direction of force.

(i) In C.G.S. system, gravitational unit of work is gram-weight-centimeter ( g wt cm).

Since W = FS

1g wt cm = 981 erg.

Thus 1g-wt-cm of work is done when a force of 1g-wt displaces a body through 1 cm in its own direction.

(ii) In S.I. system, gravitational unit of work is kilogram weight meter (kg wt m)

1kg wt m = 1kg wt × 1m = 9.81 N  × 1 m

1 kg wt m = 9.81 J

Thus, 1 kg wt m of work is done when a force of 1 kg-wt displaces a body through 1 m in its own direction.

MEASUREMENT OF WORK DONE :

By formula we know that

Work = Force × Displacement

W = F ×s

We can understand  work done by following examples:

• A man pushing a stationary wall. There is no movement of the wall. So the work done by the man on the wall is zero. However the work done on the body of the man himself is not zero. This is because when the man pushes the wall, his muscles are stretched and blood is displaced to the strained muscles more rapidly. So, due to energy consumption man feels tired.
• A man standing still at a bus stop with heavy suitcases in his hands may get tired soon but there is no work done in this situation. This is because the suitcases held by the man do not move at all.

Both the case s=0 .Distance covered by the body is zero. Then work done will be zero.

Some special cases :

Case -I : If

From equation (i)

W = Fd cos

So, W = Fd (maximum)

When force and displacement are in same direction then work done will be maximum.

Case - II : When

From equation (i)

W = Fd cos900

So, W = 0

When force and displacement are perpendicular to each other then work done will be zero.

e.g. If a body is moving in horizontal direction then work done by the force of gravity will be zero.

e.g. If a body is moving on a circular path then work done by the centripetal force will be zero, because the direction of centripetal force is towards the center of the circle and displacement will be along the tangent.

Case III :

If

then from equation (i)

W = Fd  cos 1800

then W = - Fd

When the force and displacement are in opposite direction then work done will be negative.

e.g. When a spring is compressed then the force applied by the spring and the displacement will be in opposite direction to each other, so work done by the spring will be negative.

When the spring is stretched then the work done will also be negative.

e.g. When a body of mass m in lifted upward a fore F = mg has to be applied upward 

Work done by the force of gravity will be negative

Work done, W = - mgh

Things to remember :

If F = 0 then work done, W = 0

e.g. A student revising his notes by memory without moving his limbs is doing no physical work.

A meditating saint is doing no physical work though he keeps sitting for hours.

(ii) If displacement, d = 0 then work done, W = 0.

e.g. A foolish labour trying to displace a building has done no work though he may spend the whole day.

  Erg and joule are the absolute units of work done

WORK DONE UNDER DIFFERENT SITUATIONS:

Positive Work:Positive Work done :

When the angle between force and the displacement is acute (

• In lifting a weight upward by applying an upward force, the work done by the applied force will be positive.
• In stretching a spring, the work done by the eternal force will be positive.

Negative Work:

Now consider another situation in which an object is being displaced by the action of forces and we identify one of the forces, F acting at an angle θto the direction of the displacement d, When the angle between the force and the displacement is obtuse, (

Zero Work

Thus, work done is zero when a force acts at right angles to the direction of motion of the body.

CASE – I:

When force acts at right angles to the direction of motion of a body then the angle θ between the direction of motion and direction of force is 90°.

Now, cos 90° = 0, so the component of force, F cos 90° becomes zero and hence work done = 0

W = F cos 90° ×s

W = F× 0 × s

W = 0

In this case, there in a constant force but displacement is zero, such that W = F × O = 0

WORK DONY BY THE GRAVITATIONAL FORCE:

Consider a body of mass m near the surface of the earth. Let the body be at a height h from the ground. When it is dropped, the body falls to the ground. The force that makes the body fall to the ground is the gravitational force of the earth. Since the gravitational force has made the body move thrugh a distance h, work has been done by the gravitational force.’

(i) Work Done When a Body Falls:

The gravitational acting on the body = – mg (the negative sign indicates that the gravitational force acts downwards, that is, in the –y direction.)

Work done = W = Force × distance.

  = – mg × (–h) = mgh

The work done when a body falls under graivty is positive.

(ii) Work Done When a Body is Raised against Gravity:

The weight of the body = –mg

The negative sign indicates that the weight, which is a force, acts downwards, along the –y direction. In order to lift the mass, the force applied upwards must overcome the gravitational force, the weight of the body.

The force applied = mg

Work done = Force × distance moved

W = mg × h = mgh

The work done by the external force in lifting the body is positive.

Work done in a circular path:

To keep a body moving in a circle, there is a force acting on it which is directed towards the centre. This force is called as centripetal force.

The work done on a body moving in a circular path is also zero.

This is because when a body moves in a circular path then the centripetal force acts along the radius of the circle and it is at right angle to the motion of the body.

Thus, the work done in the case of earth moving round the sun is zero and the work done in the case of a satellite moving round the earth is also zero. 

Question 1: A gatekeeper lifts a luggage of 25 kg from the ground and put it on his head, 3 m above the ground. Calculate the work done by him on the luggage. (take g = 10 m/s2.)

Solution: Mass of luggage, m = 25 kg

displacement, d = 3 m

acceleration due to gravity, g = 10 m/s2

work done, W = Fd = mgd

W = 25 × 10 × 3 = 750 J

Question 2 A force of 20 N displaces a body by 10m, the angle between force and displacement is 60o, then find the work done.

Solution: Force, F = 20 N,

displacement, d = 10m,

angle between force and displacement,

work done, W = Fd cos

then,W = 20 × 10 ×

Question 3. Calculate the work done in pushing a cart through a distance of 50 m against the force of friction which is equal to 125 N.

Solution:    F = 125 N

s = 50 m

W = F×s

W = 125 × 50 = 6,250 J.