Question

The kinetic energy of an object of mass 'm' moving with a velocity of 5 m/s is 25 J. what will be its kinetic energy when its velocity is increased three times.​

Answer 1

we know KE = (1/)mv²

so when when the velocity gets doubled KE becomes 4 times the initial KE

so KE after increasing its velocity = 4 x 25 J = 100 J

HOPE THIS HELPS YOU ✌️✌️☘️☘️. !!

Answer 2

Answer :-

$\begin{gathered}\:\\\large{\boxed{\sf{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}\end{gathered}$

Here the concept ot Kinetic Energy has been used. We see that Kinetic Energy is the half of product of mass and velocity square. Firstly, we can find the mass of the from the equation of Kinetic Energy and then we can can apply that into the new equation where there is thrice of kinetic energy.

Let's do it !!

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★ Formula Used :-

$\begin{gathered}\:\\\large{\boxed{\sf{i.)\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:v^{2}}}}}\end{gathered}$

$\begin{gathered}\:\\\large{\boxed{\sf{ii.)\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:(3\:\times\:v)^{2}}}}}\end{gathered}$

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★ Question :-

The kinetic energy of an object of mass 'm' moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is increased three times.

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★ Solution :-

Given,

» Kinetic Energy of the body = 25 Joules

» Velocity of the body = 5 m/sec

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~ For the mass of the body :-

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:v^{2}}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;25\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:(5)^{2}}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;25\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:25}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;m\;=\;\bf{\dfrac{2}{1}\:\times\:\dfrac{25}{25}\:\;=\:\;\underline{\underline{2\;\;Kg}}}}}\end{gathered}$

$\begin{gathered}\:\\\large{\boxed{\boxed{\tt{Hence,\;\;mass\;\;of\;\;the\;\;object\;\;is\;\;\boxed{\bf{2\;\;Kg}}}}}}\end{gathered}$

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~ For the Kinetic Energy of the body when velocity is three times :-

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:m\:\times\:3\:\times\:v^{2}}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:2\:\times\:(3\:\times\:5)^{2}}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:2\:\times\:(15)^{2}}}}\end{gathered}$

$\begin{gathered}\:\\\qquad \large{\sf{:\longrightarrow\;\;\;Kinetic\;Energy\;=\;\bf{1\times\:225\;\;=\;\;\underline{\underline{225\;\;Joules}}}}}\end{gathered}$

$\begin{gathered}\:\\\large{\underline{\underline{\rm{Thus,\;kinetic\;energy\;of\;the\;body\;when\;velocity\;is\;3\;times\;is\;\;\boxed{\bf{225\;\;Joules}}}}}}\end{gathered}$

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★ More formulas to know :-

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Potential\;Energy\;=\;mgh}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Heat\;Energy\;=\;Force\:\times\:Distance}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Power\;=\;\:\dfrac{Work}{Time}}\end{gathered}!/tex]</p><p> </p><p> </p><p> </p><p></p><p>[tex]\begin{gathered}\:\\\sf{\leadsto\;\;\;Pressure\;=\;\:\dfrac{Force}{Area}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Stress\;=\;\:\dfrac{Force}{Area}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Strain\;=\;\:\dfrac{Change\:in\:Dimension}{Original\:Dimension}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Coefficient\;of\;Elasticity\;=\;\:\dfrac{Stress}{Strain}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Impulse\;of\;Force\;=\;Force\:\times\:Time}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Coefficient\;of\;Viscosity\;=\;\dfrac{Force\:\times\:Distance}{Area\:\times\:Velocity}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Pressure\;Gradient\;=\;\:\dfrac{Pressure}{Distance}}\end{gathered}[tex]</p><p> </p><p> </p><p> </p><p></p><p>[tex]\begin{gathered}\:\\\sf{\leadsto\;\;\;Force\;Constant\;=\;\:\dfrac{Force}{Displacement}}\end{gathered}$

$\begin{gathered}\:\\\sf{\leadsto\;\;\;Velocity\;Gradient\;=\;\:\dfrac{Velocity}{Distance}}\end{gathered}$