Work is basically the constant force that is done on a system and it is the product of the component of the force in the direction of motion times the distance through which the force acts. And the capacity to work is called energy. Work is the term that is used for the displacement done by any force in physics. In other words, we can say that work and energy are the two essential elements to understand any physical movement.
Well, here we will discuss the work-energy theorem, limitations, and work-energy theorem examples. To complete the work, energy is needed, and hence in this theorem, we will know the relation between energy and work. Work relates to displacement, and displacement relates to kinetic energy. The work-energy theorem affirms that the work done on any object is comparable to the difference in kinetic energy of the object. So, according to the theorem statement, we can define the work-energy theorem as follows.
Kf – Ki = W
Where,
Kf = Final kinetic energy
Ki = Initial kinetic energy
W = Net-work done on the object.
Types of Work-Energy Theorem
Various types of energy are there like potential energy, kinetic energy, thermal energy, electrical energy, chemical energy, and nuclear energy. Kinetic energy is basically the type of energy that an object or particle has as an outcome of its movement and when the work is done on an object by applying net force, the item accelerates and gains kinetic energy. So the Kinetic energy is a property of a moving object or particle that is definitely by its mass as well as its motion. Now the Work-energy theorem states that the work done on a particle by the sum of all forces acting is equal to the change in the kinetic energy of the particle, in other words, net-work done on a system is equal to the change in Kinetic energy.
As we discussed earlier, kinetic energy changes due to external forces or energies like gravity or friction. You probably remember the law of energy conservation. According to the law of conservation, energy is only changed from one form to another. Let’s study the following work-energy theorem example for better understanding.
According to the Work energy theorem,
Work (completed by all kinds of energy or forces) = Change (Difference) in Kinetic Energy
Formula for Work-Energy Theorem
Wnet = K−K0 =ΔK
Wnet is net-work and K is final kinetic energy and K0 is the initial kinetic energy, and lastly, ΔK is changes in kinetic energy
Or
Wg + WN + Wf =Kf – Ki
Where Wg = work done by gravity
WN = work done by a reasonable force
Wf = work done by friction
Kf = final kinetic energy
Ki = initial kinetic energy
Work Done by a Constant Force
The constant force will result in constant acceleration. Let’s take the constant acceleration as ‘a.’
From the motion equation, we get
v2 = u2 + 2as…….. equation (1)
Therefore,
2as = v2 – u2 ………… equation (2)
In equation 2, multiply both the sides with ‘m’ mass.
(ma).s = (mv2–mu2)2
Mass X Acceleration = Force
F.s = (mv2–mu2)2…………….. equation (3)
Comparing the above equations (2) and (3), we get,
Work done by force (F) = F.s
Where’s’ is the displacement of the body.
Work is Done by (Non-Uniform) Variable Force.
Now, consider the resulting equation of work.
W = F.ds
The above equation is valid only for such conditions where the force is constant and can’t be applicable for non-constant variables.
Work-Energy Theorem for Variable Force
In the given figure, force is constant for a little displacement, and then after the force is variable till the final position. So, you can see the change in the applied force from the graph.
W = [int x_{f} x_{i} F(x)dx]
The shaded portion represents the work done by force F(x).
It represents the work that is done by a variable force. A graphical approach to this would be finding the area between F(x) and x from xi to xf.
Application of Work-Energy Theorem
The application of the Work-Energy theorem is that it is very useful in analyzing situations where a rigid body should move under several forces. A rigid body cannot store the potential energy in its lattice due to its rigid structure and it can only possess kinetic energy. The work done by any force acting on a rigid body is eventually equal to the change in its kinetic energy and it is the basis of the work-energy equation for rigid bodies.
Limitations of Work-Energy Theorem
Initially, the rule came from Newton’s second law, and hence it is applicable to the particles. Objects that are like particles are considered for this rule. So, if all the object particles behave like particles, we can consider the whole object as a particle.
Work energy theorem is used to solve different types of problems but it has a limitation that it does not give complete information about the real cause of motion which is the dynamics of Newton’s second law of motion and is called the scalar form of Newton’s second law of motion. Work energy theorem also does not define the direction of velocity.
Solved Examples
Example 1. Consider a bullet that has 20g mass and a velocity of 500 m/s. It suddenly struck a tree and went to another side with 400 m/s velocities. Find the work that is done by the bullet.
A) 900 J B) 500 J C) 800 J D) 950 J
Solution:
Mass of the bullet (given), m = 20 g (= 0.02 kg).
Beginning velocity of the bullet = 500 m/s.
Closing velocity of the bullet = 400 m/s.
Calculate the given bullet’s energy difference, and we can get the work done on the tree by the bullet.
Hence, according to the Work-Energy Theorem,
we have:
Δ(K.E.) of the bullet = 1/2{0.02(500)2 – 0.02(400)2}
Therefore,
Δ(K.E.) of the bullet = 900 J
Example 2. A block of mass 10 kg starts moving up the incline with 20m/s. It reaches the top and comes back to its initial position and stops. What is the work done by friction in the whole process?
Solution:
If we want to use the formula of work, we need the friction coefficient to calculate the frictional force. But that is not given. So let’s attempt to implement the work-energy theorem. Forces acting on the block are gravity, normal reaction, and frictional force. So,
Wf + WN + Wg = Kf − Ki
Here, WN is zero as force is always perpendicular to the displacement. Being a conservative force Wg is zero as the body returns to its initial position. Also,
Ki = 12 × 10 × 400 = 2000 J
Kf = 0
Wf = −2000 J
Hence the work done by friction is negative.
Conclusion
The article is very helpful for students to understand the work-energy theorem. It also contains some solved examples and applications of the work-energy theorem.