Question

A truck of mass 1000kg increases its speed from 36km/hr to 72km/hr. Find the increase in its kinetic energy ​

Answer 1

Answer :

➥ The increase in it's kinetic energy = 150000 J

Given :

➤ Mass of a truck (m) = 1000 kg

➤ Intial velocity of a truck (u) = 36 km/h

➤ Final velocity of a truck (v) = 72 km/h

To Find :

➤ Increase in it's kinetic energy (K.E) = ?

Solution :

Let ,

The Intial kinetic energy be "K.E₁" and the final kinetic energy be "K.E₂"

◈ Intial velocity (u) = 36 km/h = 36 × 5/18 = 10 m/s

◈ Final velocity (v) = 72 km/h = 72 × 5/18 = 20 m/s

$\tt{: \implies K.E = K.E_{1} - K.E_{2}}$

$\tt{: \implies K.E = \dfrac{1}{2}m {v}^{2} - \dfrac{1}{2}m {u}^{2} }$

$\tt{: \implies K.E = \dfrac{1}{2} \times m( {v}^{2} - {u}^{2}) }$

$\tt{: \implies K.E = \dfrac{1}{2} \times 1000( {20}^{2} - {10}^{2}) }$

$\tt{: \implies K.E = \dfrac{1}{\cancel{\:2\:}} \times \cancel{1000}(400 - 100) }$

$\tt{: \implies K.E = 1 \times 500 \times 300}$

$\tt{: \implies K.E = 500 \times 300}$

$\bf{: \implies \underline{ \: \: \underline{ \purple{ \: \: K.E = 150000 \: J\: \: }} \: \: }}$

Hence, the increase in it's kinetic energy is 150000 J.

Answer 2

$\star\:\:\:\bf\large\underline\red{Given:–}$

☛︎ Mass of the truck, m = 1000 kg

☛︎ Initial velocity, u = 36 km/hr

= 36 × 5/18

= 10 m/s

☛︎ Increased velocity, v = 72 km/hr

= 72 × 5/18

= 20 m/s

$\star\:\:\:\bf\large\underline\red{To\:find:—}$

The increase in its kinetic energy = ?

$\star\:\:\:\bf\large\underline\red{Solution:—}$

$\boxed{\bf{\blue{Formula: Kinetic\:energy=\frac{1}{2}mv²}}}$

$Initial\:kinetic\:energy=\frac{1}{2}×1000×(10)²$

$➝initial\:kinetic\:energy=50000\:J$

$Increased\:kinetic\:energy=\frac{1}{2}×1000×(20)²$

$➝Increased\:kinetic\:energy=200000\:J$

Work done = change in kinetic energy

= 200000 - 50000 J

= 150000 J

Therefore,the increase in its kinetic energy is 150000 J

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