Question

$\large{\underline{\underline{\maltese{\pink{\pmb{\sf{ \; Question \; :- }}}}}}}$

What is the work to be done 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 .

$\\ {\orange{\underline{\rule{75pt}{9pt}}}}{\color{cyan}{\underline{\rule{75pt}{9pt}}}}{\red{\underline{\rule{75pt}{9pt}}}}$

$\large{\underline{\underline{\maltese{\color{darkblue}{\pmb{\sf{ \; Note \; :- }}}}}}}$

$\longrightarrow \; {\red{\sf{ No \; Spams }}}$

$\longrightarrow$ Proper and with full Explanation .

$\longrightarrow$ Latex must be used .

$\\ {\color{cyan}{\underline{\rule{75pt}{9pt}}}}{\pink{\underline{\rule{75pt}{9pt}}}}{\color{maroon}{\underline{\rule{75pt}{9pt}}}}$

​

Answer 1

$\mathfrak\red{Question:--}$

What is the work to be done 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 .

⠀⠀⠀⠀

⠀

$\mathfrak\red{Answer:--}$

${\underline{\boxed{\bf\green{Given,}}}}$

Mass of the car, m = 1500 kg

Initial velocity of the car, u = 30 km/h =

$\bf{ \frac{30 \times 1000 \: m}{3000 \: s} = \frac{25}{3}m {s}^{ - 1} }$

[converted km/h to m/s]

⠀

Final velocity of the car, v = 60 km/h =

$\bf{ \frac{60 \times 1000 \: m}{3600 \: s} = \frac{50}{3}m {s}^{ - 1} }$

[converted km/h to m/s]

⠀⠀⠀⠀

$\\ {\orange{\underline{\rule{75pt}{9pt}}}}{\color{cyan}{\underline{\rule{75pt}{9pt}}}}{\red{\underline{\rule{75pt}{9pt}}}}$

⠀⠀⠀⠀

${\underline{\boxed{\bf\green{To \: find,}}}} - - \bf\red{work \: done \: (w)}$

⠀

⠀

${\underline{\boxed{\bf\green{Solution :}}}}$

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

=

$\bf\pink{Final \: K.E \: -Initial\: K.E }$

$\bf{work \: done \: \: w = \frac{1}{2} m {v}^{2} = \frac{1}{2} m {u}^{2} }$

$\bf{.:K.E \: = \frac{1}{m {v}^{2} } }$

where,

mass of the body = m

and the velocity with which the body is travelling = v

⠀

$➼ \: \bf\purple{W \: = \frac{1}{2}m( {v}^{2} - {u}^{2}) }$

(taking out common)

⠀

Now, Substituting the values,

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

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

⠀

$➼ \: \: \bf\purple{ {a}^{2} - {b}^{2} = (a + b)(a - b) }$

⠀

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

$\bf\red{W =156250 \: \: joule}$

⠀

➼ Hence, the work to be done to increase the velocity of a car 30 km/h-¹ to 60 km/h-¹ from is 156250 joule, if the mass of the car is 1500 kg.

⠀

⠀

$\\ {\color{cyan}{\underline{\rule{75pt}{9pt}}}}{\pink{\underline{\rule{75pt}{9pt}}}}{\color{maroon}{\underline{\rule{75pt}{9pt}}}}$

Answer 2

✏ Coversion from km/h to m/s

1 km = 1000m

1 hour = 60 minutes = 3600 s

∴ $\mathtt{\dfrac{1 \: Km}{h} = \dfrac{1000m}{3600s}}$

∴ $\mathtt{1 km/h = \dfrac{5}{18}m/s}$

✠ Solution -

→ Mass of the car (m) = 1500 kg

→ Initial Velocity (u) = 30 km/h

$\mathtt{ = 30 \times \dfrac{5}{18} \: m/s}$

$\mathtt{ = \dfrac{5 \times 5}{3} \: m/s}$

$\mathtt{ = \dfrac{25}{3} \: m/s}$

→ Finally Velocity (v) = 60 km/h

$\mathtt{ = 60 \times \dfrac{5}{18} \: m/s}$

$\mathtt{ = \dfrac{10 \times 5}{3} \: m/s}$

$\mathtt{ = \dfrac{50}{3} \: m/s}$

✠We know that,

☞ $\boxed{\small \mathtt{Work \: done = Change \: in \: kinetic \: energy}}$

So, finding initial and final kinetic energy,

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

➸ Initial Kinetic Energy

$\mathtt{ = \dfrac{1}{2}m{u}^{2}}$

$\mathtt{ = \dfrac{1}{2} \times (1500) \times \dfrac{25}{3}} \times \dfrac{25}{3}$

$\mathtt{ = \dfrac{750 \times 625}{9}}$

$\mathtt{ = \dfrac{250 \times 625}{3}}$

$\mathtt{ = \dfrac{156250}{3}J}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

➸ Final Kinetic Energy

$\mathtt{ = \dfrac{1}{2}m{v}^{2}}$

$\mathtt{ = \dfrac{1}{2} \times (1500) \times \dfrac{50}{3}} \times \dfrac{50}{3}$

$\mathtt{ = \dfrac{750 \times 2500}{9}}$

$\mathtt{ = \dfrac{250 \times 2500}{3}}$

$\mathtt{ = \dfrac{625000}{3}J}$

➸ We know that,

☞ $\boxed{\small \mathtt{Work \: done = Change \: in \: kinetic \: energy}}$

= Final Kinetic Energy - Initial Kinetic Energy

$\mathtt{ = \dfrac{625000}{3}J - \dfrac{156250}{3}J}$

$\mathtt{ = \dfrac{468750}{3}J}$

$= \boxed{\mathtt{\dfrac{156250}{3}J}}$

⇝ Work done is 156250 Joules.