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 of 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.

As we discussed earlier, the 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.

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Work energy theorem class 9

According to Work energy theorem,

Work (completed by all kind of energy or forces) = Change (Difference) in Kinetic Energy

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

The constant force will result in constant acceleration. Let's take the constant acceleration as 'a.'

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Work done by a constant force

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.

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.

In the above 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 = ∫xfxiF(x)dx

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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 .

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.

Example 1

Consider a bullet that has 20g mass and velocity 500 m/s. It suddenly struck a tree and went to another side with 400 m/s velocity. 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.

FAQ (Frequently Asked Questions)

1. Steps to Approach Problems on Work Energy Theorem?

Problems of work energy theorems seem to be difficult but should be solved with proper steps to get accurate results. Well, your logic and thinking skill help you here to get it answered quickly through a method -oriented approach. Here are some necessary steps to be considered to solve the problems and strategy should be adopted accordingly.

**Following Steps Should be Considered:**

**Step-1: **Map the FBD of the object, thus recognising the forces operating on the purpose.

**Step-2:** Getting the initial and concluding kinetic energy.

**Step-3:** Associating the values according to the theorem.

2. How Can We Efficiently Use This Theorem?

Problems related to work and applied force can be solved with this theorem, or you can calculate both the entities separately. Work energy theorem gives an accurate conclusion avoiding lengthy process. If you work on separate values of both work and force, it won't be easy to calculate. Work energy theorem helps understand the relationship between work and energy (kinetic and potential) in a particular case. As we know that energy to do any task may differ according to the situation, and hence through this theorem, you can get standard information about work and applied energy.