Table of contents  
Work  
Not Much ‘work’ In Spite Of Working Hard!  
Scientific Conception of Work  
Work Done by a Constant Force  
Units of Work  
Some Solved Examples 
In our daytoday lives, we often use the term 'work' to describe both mental and physical efforts. However, the concept of work in physics is distinctly different from this everyday understanding. To make this point clear let us consider a few examples.
Examples:
In all these cases, according to scientific definition, you are not doing any work. Let's dive into the fascinating world of physics and explore this unique perspective on work!
In physics, the term work is used in a special technical sense and has a much more precise definition.
In physics, work is the measure of energy transfer that occurs when an object is moved by an external force along a displacement. It is defined as the product of the force acting on the object and the distance it is moved in the direction of the force.
Examples:
From all the examples given above, it follows that work is done if:
Note: No work is said to be done if any of the two conditions is not satisfied.
When Constant Force is Acting in the Direction of Displacement
When a Constant Force is Acting at an Angle to the Displacement
Case1:
Case2:
Work done in Various Situations
Case3:
When the angle between force and displacement is 180°,
i.e. when θ =180° then cos 180° = 1.
In this case, work done is negative.
Example: A block A is pushed by the force F and displaced through a distance s.
Work Done by Friction is Negative
In CGS system:
In SI system:
Definition of One Joule
When F = 1N, s = 1m, then W = 1J.
If a displacement of 1m is produced by a force of 1N acting in the direction of displacement then the work done by the force is 1J.
1J= 1 N x 1 m = 105 dyne × 102 cm = 107 erg
Example 1:
Imagine you are pushing a 10 kg box across a smooth, flat floor. You are applying a force of 20 N at an angle of 30° above the horizontal direction. The box moves 5 meters to the right. Calculate the work done on the box.
Solution: In this situation, the displacement is in the horizontal direction (5 meters to the right), and the force has a horizontal component as well as a vertical component. To find the work done, we only need to consider the horizontal component of the force, which is given by Fx = F * cos θ, where F is the applied force and θ is the angle between the force and the displacement.
Fx = 20 N * cos(30°) = 20 N * (√3 / 2) = 10√3 N
Now, we can calculate the work done using the formula W = Fx * d, where d is the displacement.
W = 10√3 N * 5 m = 50√3 J
Thus, the work done on the box is 50√3 Joules. Since the angle is acute (30°), the work done is positive, and since the force has a component in the direction of the displacement, the work done is maximum.
Example 2:
A coolie carries a load of 50 kg on his head and walks along a horizontal surface. He applies a force on the load in the vertically upward direction. He walks a distance of 100 meters. Calculate the work done by the coolie on the load.
Solution:
Given,
Mass of load = 50 kg
Distance traveled = 100 m
The force applied by the coolie is in the vertically upward direction, and the displacement of the load is in the horizontal direction. Therefore, the angle between the force and the displacement, θ = 90°.
We know that work done, W = F × d × cos θ
Since θ = 90°, cos 90° = 0.
Hence, work done, W = F × d × 0 = 0
Thus, the work done by the coolie on the load is 0 Joules.
Example 3:
A person pushes a 10 kg block with a force of 50 N along a horizontal surface for a distance of 4 meters. The frictional force acting on the block is 30 N. Calculate the work done by the applied force and the work done by the frictional force.
Solution:
Given:
Mass of the block, m = 10 kg
Applied force, F = 50 N
Distance, s = 4 m
Frictional force, f = 30 N
First, let's find the work done by the applied force:
W_{applied} = F × s × cos 0°
W_{applied} = 50 N × 4 m × 1 (since cos 0° = 1)
W_{applied} = 200 J (Joules)
Now, let's find the work done by the frictional force:
W_{friction} = f × s × cos 180°
W_{friction} = 30 N × 4 m × (1) (since cos 180° = 1)
W_{friction} = 120 J (Joules)
So, the work done by the applied force is 200 J, and the work done by the frictional force is 120 J.
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