Question

What is the work to be done to increase the velocity of a car from 36km/hr
to 72 km/hr if the mass of the car is 2000Kg?​

Answer 1

GIVEN :-

• Velocity of car is increased by 36km/hr to 72km/hr.
• Mass of car is 2000kg.

TO FIND :-

• Work done.

TO KNOW :-

Work done is change is kinetic energy.

$\\ \boxed{\sf \: W = \Delta K.E}$

Also ,

$\\ \boxed{\sf \: K.E = \dfrac{1}{2} m {v}^{2} }$

Here ,

• W → Work done
• K.E → Kinetic energy
• m → Mass of object
• v → Velocity of object

SOLUTION :-

Converting factor is (5/18).

We have ,

• Initial velocity (u) = 36km/hr

We will convert it into m/s.

Initial velocity = (5/18) × 36

Initial velocity = 10m/s

• Final velocity (v) = 72km/hr

We will convert it into m/s.

Final velocity = (5/18) × 72

Final velocity = 20m/s

Mass of object (m) = 2000kg

$\\ \sf \: </strong><strong>W</strong><strong> = </strong><strong>F</strong><strong>inal \: </strong><strong>K.</strong><strong>E</strong><strong> - </strong><strong>I</strong><strong>nitial \: </strong><strong>K.E</strong><strong> \\ \\ </strong><strong>\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong> \sf \: </strong><strong>W</strong><strong> = \</strong><strong>d</strong><strong>frac{1}{2} m {v}^{2} - \</strong><strong>d</strong><strong>frac{1}{2} m {u}^{2} \\ \\ \\ </strong><strong>\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong> \sf \: </strong><strong>W</strong><strong> = \</strong><strong>d</strong><strong>frac{1}{2} m( {v}^{2} - {u}^{2} )$

Putting values we get,

$\\ \implies\sf \: W = \dfrac{1}{ \cancel2} ( \cancel{2000}) \{( {20)}^{2} - {(10)}^{2} \} \\ \\ \\ \implies \sf \: W = 1000(400 - 100) \\ \\ \\ \implies \sf \: W = 1000(300) \\ \\ \\ \implies\underbrace{\boxed{\sf \: W = 300000J} }$

Hence , 300000J or 300kJ of work is done .