Question

13. Mass of a car is 1000 kg. If its speed increase from
36 km/hr to 72km/hr then find gain in its kinetic energy.
Ans. 1.5 x 1053

Answer 1

$\rm Answer$

$\rm Given -$

$\bf m = 1000 kg$

$\bf u = 36 km/hr$

$\bf v = 72 km/hr$

where

$\longrightarrow$m is mass of car.

$\longrightarrow$u is initial velocity.

$\longrightarrow$v is final velocity.

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$\rm To \:find -$

Gain in kinetic energy

━━━━━━━━━━━━

$\rm Formula\: used -$

$\boxed{\bf\pink{ Kinetic \:energy = \frac{1}{2} mv^{2}}}$

where

$\longrightarrow$v is velocity.

$\longrightarrow$m is mass of object.

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$\rm Solution -$

Kinetic energy in 1st case -

$\longrightarrow$$\bf m = 1000 kg$

$\longrightarrow$$\bf u = 36 km/hr = 10 m/s$

$\bf Kinetic\: energy = \frac{1}{2} \times m \times {u}^{2}$

$\implies$$\bf \frac{1}{2} \times 1000 \times 10^2$

$\implies$$\bf\purple{ 50000 kg {m}^{2}/{s}^{2}}$

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Kinetic energy in 2nd case -

$\longrightarrow$$\bf m = 1000 kg$

$\longrightarrow$$\bf v = 72 km/hr = 20 m/s$

$\bf Kinetic\: energy = \frac{1}{2} \times m \times {v}^{2}$

$\implies$$\bf \frac{1}{2} \times 1000 \times 20^2$

$\implies$$\bf\purple{ 200000 kg {m}^{2}/{s}^{2} }$

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Gain in kinetic energy

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

$\implies$$\bf 200000 - 50000$

$\implies$$\bf 150000 kg {m}^{2}/{s}^{2}$

$\implies$$\bf 1.5 \times 10^5 kg {m}^{2}/{s}^{2}$

$\sf\red{Gain\: in \:kinetic\: energy = 1.5 \times 10^5 kg {m}^{2}/{s}^{2}}$

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Answer 2

$\huge{\underline{\mathbb{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

$\large\underline\bold\red{Given}$

• $\bf m = 1000 kg$

• $\bf u = 36 km/hr$

• $\bf v = 72 km/hr$

where : -

• m $\dashrightarrow$mass of car.

• u $\dashrightarrow$ initial velocity.

• v $\dashrightarrow$final velocity.

$\large\underline\bold\green{To\: find}$

• Gain in kinetic energy

$\large\underline\bold\red{formula\: we \:use}$

$\implies$$\bf Kinetic \:energy = \frac{1}{2} mv^{2}$

where

• v $\dashrightarrow$ velocity.

•m $\dashrightarrow$mass of object.

$\large\underline\bold\red{kinetic\: Energy\:in \:first \: case}$

• $\bf m = 1000 kg$

• $\bf u = 36 km/hr = 10 m/s$

$\implies$$\bf Kinetic\: energy = \frac{1}{2} \times m \times {u}^{2}$

$\implies$$\bf \frac{1}{2} \times 1000 \times 10^2$

$\implies$$\bf 50000 kg {m}^{2}/{s}^{2}$

$\large\underline\bold\red{kinetic\:Energy\: in\:second \: case }$

$\implies$$\bf m = 1000 kg$

$\implies$$\bf v = 72 km/hr = 20 m/s$

$\bf Kinetic\: energy = \frac{1}{2} \times m \times {v}^{2}$

$\bf \frac{1}{2} \times 1000 \times 20^2$

$\implies$$\bf 200000 kg {m}^{2}/{s}^{2}$

$\implies$Gain in kinetic energy

$\implies$$\bf = (\frac{1}{2} m {v}^{2})-(\frac{1}{2} m {u}^{2})$

$\implies$$\bf 200000 - 50000$

$\implies$$\bf 150000 kg {m}^{2}/{s}^{2}$

$\implies$$\bf 1.5 \times 10^5 kg {m}^{2}/{s}^{2}$

$\large\underline\bold\green{Verified}$