Question

kinetic energy of an object mass, m moving with a velocity of 5 ms-1 is 25 joule .what will be its kinetic energy when its velocity is doubled?what will be its kinetic energy when its velocity is increased three times?

Answer 1

Mass = m v=5 m/s K.E = 25 joule K.E =1/2 mv^2 25= m/2× 25 m= 1 × 2 = 2 joule When V= 2 v = 10 m/s K.E = 1/2 mv^2       = 1/2×2×10× 10       = 100 So kinetic energy will become 4 times

Answer 2

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Here, in this given query in first case , we have to find out the mass,m by applying the formula of kinetic energy .

In the second case , we have to find out the kinetic energy by applying the formula of kinetic energy . Note - As given the velocity will be doubled. Suppose if the given velocity is 5 then if it is doubled then the new velocity will be

• 5 × 2 = 10

In the third case , we have to find out the kinetic energy by applying the formula of kinetic energy . Note - As given the velocity will be tripled . Suppose if the given velocity is 5 then if it is tripled then the new velocity will be

• 5 × 3 = 15

$\star \bigstar \rm { \underbrace{\underline{ SoLuTioN }}}\bigstar \star$

$\rm \: Given,$

$\rm ⏣ \: Initial \: velocity, v_{1} \longrightarrow \: 5 \: m/s$

$\rm ⏣ \: Kinetic \: energy, E _{k _{1} }\longrightarrow \: 25 \: J$

$\rm ⏣ \: Mass,m \longrightarrow ?$

$\rm \: We\: know,$

$\rm \boxed{\: \rm \red E_{k _{1} }= \frac{1}{2} mv _{1} {}^{2} }$

$\rm \implies25 = \frac{1}{2} \times m \times 5 \times 5$

$\rm \implies25 = \frac{25 \: m}{2}$

$\rm \implies 50=25 \: m$

$\rm \implies 25 \: m=50$

$\rm \implies m= \cancel\frac{50 {}^{2} }{25}$

$\rm \implies m=2 \: kg$

$\rm \: When \: velocity \: is \: doubled,then$

$\rm \: ⏣ \: v_{2} \longrightarrow \: 10 \: m/s$

$\rm ⏣ \: mass,m \longrightarrow \: 2 \: kg$

$\rm \: We \: know,$

$\rm \boxed{\: \rm \red E_{k _{2} }= \frac{1}{2} mv _{2} {}^{2} }$

$\rm \implies\rm E_{k _{2} }= \frac{1}{ \cancel2} \times \cancel2 \times 10 \times 10 \: J$

$\rm \implies \rm E_{k _{2} } =100 \: J\:$

$\rm \: When \: velocity \: is \: tripled,then$

$\rm ⏣ \: v_{3} \longrightarrow \: 15 \: m/s$

$\rm ⏣ \: Mass,m \longrightarrow \: 2 \: kg$

$\rm ⏣ \: Kinetic \: energy, E_{k _{3}} \longrightarrow \: ?$

$\rm We\: know,$

$\rm \boxed{\: \rm \red E_{k _{3}}= \frac{1}{2} mv _{3} {}^{2} }$

$\rm \implies E_{k _{3}}= \frac{1}{ \cancel2} \times \cancel 2 \times 15 \times 15 \: J$

$\rm \implies \: E_{k _{3}}= 225 \: J$

$\\ \bigstar{ \underline{ \underline \pink{ \sf★@iTzShInNy☆}}} \bigstar$