Question

if mass of a moving body is doubled how many times will the kinetic energy increase???​

Answer 1

Given:-

• The mass of a moving body is doubled.

To Find:-

• How many times will the kinetic energy increase ?.

Formulae used:-

• Kinetic energy = ½ × mv²

Where,

• m = Mass
• v = Velocity

Now,

→ Old K. E = ½ × mv²......eq. 1

Atq, if the mass of moving body is Doubled,

→ i.e, = m = 2m

Therefore,

→ K. E' = ½ × 2m × v²

→ K. E' = 2 ( ½ × m × v² )....eq. 2

Now, Putting the value of eq. 1 in eq. 2

→ K. E' = 2 ( K. E)

Hence, The new kinetic energy will increase by 2 times.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

If mass of a moving body is doubled how many times will the kinetic energy increase ?

$\huge{\boxed{\rm{\red{Answer}}}}$

$\large{\boxed{\sf{Given \: that}}}$

• The mass of a moving body is doubled.

$\large{\boxed{\sf{To \: find}}}$

• How many times will the kinetic energy increase.

$\large{\boxed{\sf{Solution}}}$

• At 2 times will the kinetic energy increase.

$\large{\boxed{\sf{Full \: solution}}}$

$\large{\boxed{\sf{Formula \: used}}}$

$\large\underbrace\mathfrak\green{Kinetic \: energy \: formula}$

$\large\purple{\texttt{½ × mv²}}$

$\large\underbrace\mathfrak\green{Terms \: meaning \: in \: ½×mv²}$

□ $\large\pink{\texttt{m means mass}}$

□ $\large\pink{\texttt{v means velocity}}$

$\large\purple{\texttt{Now what to do ?}}$

$\large\purple{\texttt{Now,}}$

$\large\underbrace\mathfrak\green{K.E \: means \: velocity}$

✧ K.E = ½ × mv² Equation 1

If mass of moving body is doubled then,

{ It's mean like a² means a × a or 2a }

✷ m = 2m

$\large\purple{\texttt{Therefore,}}$

$\mapsto$ K.E = ½ × 2m × v²

$\mapsto$ K.E = 2 ( ½ × m × v² ) Equation 2

$\large\purple{\texttt{Substituting the value we get }}$

$\large\purple{\texttt{Substitute values of equation 1}}$

$\mapsto$ K.E = K.E = 2 (K.E)

$\implies$ $\large{\boxed{\boxed{\sf{2 \: is \: the \: answer}}}}$

Hence, we get our result that is 2.

Means 2 times will the kinetic energy increase.

Hope it's helpful

Thank you :)