Question

Hola brainly wizards,

Solve the question mentioned below:-

What work is said to be done to increase the velocity of a car from 15 km/h to 30 km/h, if the mass of
the car is 1000 kg?​

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1. Don't spam.
2. Answer should be relevant.

​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Kinetic Energy has been used. We are given the initial and final velocities of the body. Firstly we can convert them in standard units this is because Unit of Workdone is Joules and it is given as Kg m/sec². After this we can apply the value and find the answer.

Let's to it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Kinetic\;Energy_{(change)}\;=\;\bf{\dfrac{1}{2}\:mv^{2}\;-\;\dfrac{1}{2}\:mu^{2}}}}$

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★ Solution :-

Given,

» Initial Velocity of car, u = 15 Km/hr

» Final Velocity of car, v = 30 Km/hr

» Mass of the car, m = 1000 Kg

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~ For Standard units of Velocities ::

We see that for the unit Joule, we don't have velocity of km/hr. So first we need to change the velocities in m/sec.

Then,

• Initial Velocity, u -

$\\\;\sf{\rightarrow\;\;15\;Km\:hr^{-1}\;=\;\bf{\dfrac{15\;\times\;1000}{60\;\times\;60}\;m\:sec^{-1}}}$

$\\\;\sf{\rightarrow\;\;15\;Km\:hr^{-1}\;=\;\bf{\dfrac{15\;\times\;10}{6\;\times\;6}\;m\:sec^{-1}}}$

$\\\;\sf{\rightarrow\;\;15\;Km\:hr^{-1}\;=\;\bf{\dfrac{5\;\times\;5}{6}\;m\:sec^{-1}}}$

$\\\;\bf{\rightarrow\;\;15\;Km\:hr^{-1}\;=\;\bf{\dfrac{25}{6}\;m\:sec^{-1}}}$

• Final Velocity, v -

$\\\;\sf{\rightarrow\;\;30\;Km\:hr^{-1}\;=\;\bf{\dfrac{30\;\times\;1000}{60\;\times\;60}\;m\:sec^{-1}}}$

$\\\;\sf{\rightarrow\;\;30\;Km\:hr^{-1}\;=\;\bf{\dfrac{30\;\times\;10}{6\;\times\;6}\;m\:sec^{-1}}}$

$\\\;\sf{\rightarrow\;\;30\;Km\:hr^{-1}\;=\;\bf{\dfrac{5\;\times\;10}{6}\;m\:sec^{-1}}}$

$\\\;\bf{\rightarrow\;\;30\;Km\:hr^{-1}\;=\;\bf{\dfrac{50}{6}\;m\:sec^{-1}}}$

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~ For Workdone to increase the Velocity ::

We know that,

$\\\;\;\sf{:\Longrightarrow\;\!Kinetic\;Energy_{(\Delta)}\;=\;\bf{Final\;K.E.\;-\;Initial\;K.E.}}$

By applying values, we get,

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{1}{2}\:mv^{2}\bigg)\;-\;\bigg(\dfrac{1}{2}\:mu^{2}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{1}{2}\:\times\:1000\;\times\;\bigg(\dfrac{50}{6}\bigg)^{2}\bigg)\;-\;\bigg(\dfrac{1}{2}\:\times\:1000\;\times\;\bigg(\dfrac{25}{6}\bigg)^{2}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{1}{2}\:\times\:1000\;\times\;\dfrac{2500}{36}\bigg)\;-\;\bigg(\dfrac{1}{2}\:\times\:1000\;\times\;\dfrac{625}{36}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(500\;\times\;\dfrac{2500}{36}\bigg)\;-\;\bigg(500\;\times\;\dfrac{625}{36}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{1250000}{36}\bigg)\;-\;\bigg(\dfrac{312500}{36}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{1250000\;-\;312500}{36}\bigg)}}$

$\\\;\;\sf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{\bigg(\dfrac{937500}{36}\bigg)}}$

$\\\;\;\bf{:\Longrightarrow\;\;Kinetic\;Energy_{(\Delta)}\;=\;\bf{26041.67\;\;Joules}}$

Also, 1 Joule = 1 Kg m²/sec²

Even, this Change in Kinetic Energy, ∆ = Work Done.

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;Workdone\;=\;\bf{26041.67\;\;Joules}}}}$

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★ More to know :-

$\\\;\tt{\leadsto\;\;Workdone\;=\;Force\;\times\;Distance\;=\;[M^{1}\:L^{2}\:T^{-2}]}$

$\\\;\tt{\leadsto\;\;Pressure\;=\;\dfrac{Force}{Area}\;=\;[M^{1}\:L^{-1}\:T^{-2}]}$

$\\\;\tt{\leadsto\;\;Power\;=\;\dfrac{Work}{Time}\;=\;[M^{1}\:L^{2}\:T^{-3}]}$

$\\\;\tt{\leadsto\;\;Force\;=\;Mass\;\times\;Acceleration\;=\;[M^{1}\:L^{1}\:T^{-2}]}$

$\\\;\tt{\leadsto\;\;Potential\;Energy\;=\;Mass\;\times\;Acceleration\;\times\;Height\;=\;[M^{1}\:L^{2}\:T^{-2}]}$

Answer 2

∴ Work done =337500 Joule.

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