Question

Example 11.4 What is the work to
to increase the velocity of a car from
30 km h-1 to 60 km h-1 if the mass of
the car is 1500 kg?​

Answer 1

$\underline{\underline{\textsf{Solution :}}}$

We are provided with initial velocity (u) as 30 km/hr and final velocity (v) as 60 km/hr and we are also given the mass of car (m) = 1500 kg. And we are asked to find the work done by the car. To solve this question first we have to convert the initial velocity and final velocity from km/hr to m/s. and after converting the units to m/s we will find the work done by the car by using given below formula :

• Work done = Change in Kinetic energy

Now, we have to simply plug in the known values in the above given formula and we get our final answer. So, Let's do it :

$\underline{\bigstar \: \textsf{Converting the units from km/hr to m/s :}}$

$\dashrightarrow\:\:\sf Initial \: Velocity = 30 \: km/hr$

$\dashrightarrow\:\:\sf Initial \: Velocity = 30 \times \dfrac{5}{18} \: m/s$

$\dashrightarrow\:\: \underline{ \boxed{\sf Initial \: Velocity = \dfrac{150}{18} \: m/s}}$

$\dashrightarrow\:\:\sf Final \: Velocity = 60 \: km/hr$

$\dashrightarrow\:\:\sf Final \: Velocity = 60 \times \dfrac{5}{18} \: m/s$

$\dashrightarrow\:\: \underline{ \boxed{\sf Final \: Velocity = \dfrac{300}{18} \: m/s }}$

$\underline{\bigstar \: \textsf{According to the Question Now :}}$

$:\implies \sf Work \: done = Change \:in \:Kinetic\: energy$

$:\implies \sf Work \: done = Final_{(Kinetic\: energy)}-Initial_{(Kinetic\: energy)}$

$:\implies \sf Work \: done = \dfrac{1}{2}m {v}^{2} - \dfrac{1}{2}m {u}^{2}$

$:\implies \sf Work \: done = \dfrac{1}{2}m( {v}^{2} - {u}^{2} )$

$:\implies \sf Work \: done = \dfrac{1}{2} \times 1500 \bigg \lgroup \bigg(\dfrac{300}{18}\bigg)^{2} - \bigg( \dfrac{150}{18} \bigg)^{2} \bigg \rgroup$

$:\implies \sf Work \: done = 750 \bigg \lgroup \dfrac{90000}{324} - \dfrac{22500}{324} \bigg \rgroup$

$:\implies \sf Work \: done = 750 \bigg \lgroup \dfrac{90000 - 22500}{324} \bigg \rgroup$

$:\implies \sf Work \: done = 750 \bigg \lgroup \dfrac{67500}{324} \bigg \rgroup$

$:\implies \sf Work \: done =\dfrac{50625000}{324}$

$:\implies \underline{ \boxed{ \sf Work \: done =156250 \: J}}$

Answer 2

Answer:

Work done is 156250 J.

Explanation:

Given,

The mass (m) of the car is 1500 kg.

Initial velocity of car = u = 30 km / hr = 30 * 5/18 = 25 / 3 m/s

Final velocity of car = v = 60 km / hr = 60 * 5/18 = 50/3 m/s

To find: The work to be done to increase the velocity of a car from

30 km h-1 to 60 km h^-1

Solution:

Initial kinetic energy = Ki = 1/2mu² = 1/2*1500*25/3 * 25/3 = 156250/3 J

Final Kinetic energy = Kf = 1/2mv² = 1/2*1500*50/3*50/3 = 625000/3 J

Work done can be given as change in Kinetic Energy,

i.e W = ΔK.E

or, W = Kf -  Ki

or, W = 625000/3 - 156250/3

or, W = 156250 J

Therefore, Work done is 156250 J.