Question

A truck of mass 100 kilogram increases its speed from 36 kilometre per hour to 72 kilometre per hour find the increase is it's kinetic energy?

Answer 1

Answer:

$\boxed{\mathfrak{Increase \ in \ trucks \ kinetic \ energy = 15 \ kJ}}$

Given:

Mass of truck (m) = 100 kg

Initial speed (u) = 36 km/hr = 10 m/s

Final speed (v) = 72 km/hr = 20 m/s

Explanation:

Formula of kinetic energy:

$\boxed{ \bold{KE = \frac{1}{2} m {v}^{2} }}$

Increase in kinetic energy of truck (∆KE):

$\rm \implies \Delta KE = \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {u}^{2} \\ \\ \rm \implies \Delta KE = \dfrac{1}{2} m ({v}^{2} - {u}^{2} ) \\ \\ \rm \implies \Delta KE = \dfrac{1}{2} \times 100( {20}^{2} - {10}^{2} ) \\ \\ \rm \implies \Delta KE = 50(400 - 100) \\ \\ \rm \implies \Delta KE = 50 \times 300 \\ \\ \rm \implies \Delta KE = 15000 \: J \\ \\ \rm \implies \Delta KE = 15 \: kJ$

Answer 2

Answer:

72 kmph = 72 × 5/18 m/s = 20 m/s

36 kmph = 36 × 5/18 m/s = 10 m/s

Increase in kinetic energy = 0.5 m (v² - u²)

= 0.5 × 100 kg × [(20 m/s)² - (10 m/s)²]

= 15 × 10³ J

= 15 kJ

Increase in its kinetic energy is 15 kJ