Question

A truck of mass 1000kg, increases it's speed from 36km/hour to 72km/hour. Find the increase of it's kinetic energy.

Answer 1

ANSWER :

Given :

▪ Mass of truck = 1000kg

▪ Initial velocity of truck = 36kmph

▪ Final velocity of truck = 72kmph

To Find :

▪ Change in kinetic energy.

Formula :

✏ Formula of kinetic energy in terms of mass of body and velocity of body is given by...

☞ K = 1/2mv^2

Conversion :

✴ 1kmph = 5/18mps

✴ 36kmph = 36×5/18 = 10mps

✴ 72kmph = 72×5/18 = 20mps

Calculation :

✏ ΔK = K2 - K1

✏ ΔK = 1/2mv^2 - 1/2mu^2

✏ ΔK = 1/2m(v^2-u^2)

✏ ΔK = 1/2×1000(400-100)

✏ ΔK = 1/2×1000(300)

✏ ΔK = 1,50,000J = 150kJ

Answer 2

Given :

◾Mass of the truck = 1000 kg

◾Initial speed of truck = 36 kmh¯¹ = 10ms¯¹

◾Final speed of truck = 72 kmh¯¹ = 20ms¯¹

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To Find :

Increase in the kinetic energy of truck

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Formula used :

$Kinetic \: Energy \: (K.E) = \frac{1}{2} m {v}^{2}$

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Solution :

Let the Initial kinetic energy be K.Ei.

And, the final kinetic energy be K.Ef.

Now, according to the question

$Increase \: in \: K.E = K.Ef - K.Ei \\ </p><p>Increase \: in \: K.E = \frac{1}{2} m {(vf)}^{2} - \frac{1}{2}m {(vi)}^{2} \\ Increase \: in \: K.E = \frac{1}{2} \times 1000 \times ( {20}^{2} - {10}^{2} ) \\ Increase \: in \: K.E = \frac{1}{2} \times 1000 \times 300$

$Increase \: in \: K.E = 150,000 J$