What is the work to be done to increase the velocity of a car from 36km h to 72km hif the mass of the car is 1000 kg?

Free

10 Questions 10 Marks 6 Mins

Concept:

• Work-energy theorem: It states that work done by a force acting on a body is equal to the change in the kinetic energy of the body i.e.,

W = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{\Delta }}K\,\)

Where v = final velocity, u = initial velocity and m = mass of the body

Calculation:

Given - mass (m) = 1500 kg, initial velocity (u) = 36 km/h = 10 m/s and final velocity (v) = 72 km/h = 20 m/s

• The work done to increase the velocity of a 1500 kg car from 36 km/h to 72 km/h is

\(\Rightarrow W = \frac{1}{2}m({v^2} - {u^2} )\,\)

\(\Rightarrow W = \frac{1}{2}\times 1500\times ({400} - {100} )=2.25\times10^5\, J\,\)

Stay updated with the Physics questions & answers with Testbook. Know more about Work Power and Energy and ace the concept of Work.

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.4 Crore+ Students

Text Solution

Answer : A::B

Solution : Here, `u = 30km//h = (30xx(5)/(18))m//s = (25)/(3)m//s (1km//h = (5)/(18) m//s)` <br> `upsilon = 60km//h = (60xx(5)/(18))m//s= (50)/(3)m//s` <br> m = 1500kg <br> According to work-energy theorem, <br> `W = (1)/(2)m upsilon^2 - (1)/(2)mu^2 = (1)/(2)m(upsilon^2 - u^2)` <br> or `W =(1)/(2)xx1500kg[((50)/(3)m//s)^2 - ((25)/(3)m//s)^2]` <br> `=750kg[((2500)/(9)-(625)/(9))(m//s)^2]` <br> `=750kgxx208xx208.33(m//s)^2 = 156250J`

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Given,

Mass of the car, m = 1500 kg 

Initial velocity of the car, u = 30 km/h = $\frac{30\times 1000m}{3600s}=\frac{25}{3}m/s$              [converted km/h to m/s]

Final velocity of the car, v = 60 km/h =  $\frac{60\times 1000m}{3600s}=\frac{50}{3}m/s$               [converted km/h to m/s]

To find = Work done (W)

Solution:

According to the Work-Energy theorem or the relation between Kinetic energy and Work done - the work done on an object is the change in its kinetic energy.

So, Work done on the car = Change in the kinetic energy (K.E) of the car

                                         = $Final\ K.E-Initial\ K.E$

$Work\ done, \ W =\frac{1}{2}m{v}^{2}-\frac{1}{2}m{u}^{2}$  $[\because K.E=\frac{1}{m}{v}^{2}, \ where, \ mass\ of\ the\ body=m,\ and\ the\ velocity\ with\ which\ the\ body\ is\ travelling=v]$

$W=\frac{1}{2}m[{v}^{2}-{u}^{2}]$                       $[taking\ out\ common]$

Now, substituting the values-

$W=\frac{1}{2}\times 1500[(\frac{50}{3}{)}^{2}-(\frac{25}{3}{)}^{2}]$

$W=\frac{1}{2}\times 1500[(\frac{50}{3}+\frac{25}{3})(\frac{50}{3}-\frac{25}{3})]$     $[\because ({a}^{2}-{b}^{2})=(a+b)(a-b)]$

$W=\frac{1}{2}\times 1500\times \frac{75}{3}\times \frac{25}{3}$

$W=156250J$

Hence, the work to be done to increase the velocity of a car from 30km/h to 60km/h is 156250 joule, if the mass of the car is 1500 kg.