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Question: What is the work done to increase the velocity of a car from \( 30km{{h}^{-1}} \) to \( 60km{{h}^{-1...

What is the work done to increase the velocity of a car from 30kmh−130km{{h}^{-1}} to 60kmh−160km{{h}^{-1}} if the mass of the car is 1500kg1500kg ?

Explanation

Solution

Hint
In the given question, we have been given the initial and the final velocities of the car and the mass of the car and we have been asked to find the work done to increase the velocity of the car. The work done is normally calculated as a product of the force applied on a body and the displacement of the body, but we haven’t been given the displacement of the car. We can proceed with the work-energy theorem. Let’s go through a detailed solution.
⇒W=Kf−Ki\Rightarrow W={{K}_{f}}-{{K}_{i}} , K=12mv2K=\dfrac{1}{2}m{{v}^{2}}

Complete step by step answer
As discussed above, we can use the work-energy theorem to find the work done. The work-energy theorem states that the work done on a body is equal to the change in the kinetic energy of the body. In simple terms, we’ll find the initial kinetic energy and the final kinetic energy of the car and the difference between the two will give us the work done.
The initial velocity of the car (u)=30kmh−1=30×518m/s=253m/s(u)=30km{{h}^{-1}}=30\times \dfrac{5}{18}m/s=\dfrac{25}{3}m/s
The initial kinetic energy of the car can be calculated as follows
(Ki)=12mu2=12×1500×(253)2 ⇒Ki=52083.33J \begin{aligned} & ({{K}_{i}})=\dfrac{1}{2}m{{u}^{2}}=\dfrac{1}{2}\times 1500\times {{\left( \dfrac{25}{3} \right)}^{2}} \\\ & \Rightarrow {{K}_{i}}=52083.33J \\\ \end{aligned}
The final velocity of the car (v)=60kmh−1=60×518m/s=503m/s(v)=60km{{h}^{-1}}=60\times \dfrac{5}{18}m/s=\dfrac{50}{3}m/s
The final kinetic energy of the car can be calculated as follows
(Kf)=12mv2=12×1500×(503)2 ⇒Ki=208333.33J \begin{aligned} & ({{K}_{f}})=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}\times 1500\times {{\left( \dfrac{50}{3} \right)}^{2}} \\\ & \Rightarrow {{K}_{i}}=208333.33J \\\ \end{aligned}
The work done can be calculated as W=Kf−KiW={{K}_{f}}-{{K}_{i}}
Substituting the values, we get
⇒W=(208333.33−52083.33)J=156250J\Rightarrow W=\left( 208333.33-52083.33 \right)J=156250J
This is the required work done.

Note
In some questions and cases, when work is done on a body, there is no change in the velocity or the speed of the body but the position of the body. Since the velocity remains the same, we cannot make use of the kinetic energy to find the work done. In those questions, we make the use of potential energy of the body to find the work done. The work done is given as the difference of the initial and the final potential energies.