Question

ii) What is the work to be done to increase the velocity of a car from 72 km/h to 90 km/h if the mass of the car is 1800 Kg.

Answer 1

We know, as per the work - energy theorem,

$\boxed{\sf{W = ∆KE = \frac{1}{2} m ({v_{f}}^2 - {v_{i}}^2)} }$

We are given that, initial velocity = 72 km/h or 20 m/s, final velocity = 90 km/h or 25 m/s and mass of the car is 1800 kg.

∴ W = ½(1800 kg)[(25 m/s)² - (20 m/s)²]

⇒ W = (900 kg)[(25 + 20)(25 - 20) m/s² × m]

⇒ W = (900 kg)(45 × 5 m/s² × m)

⇒ W = (900)(225) Nm

⇒ W = 202500 J or 202.5 kJ (answer).

More:-

Formula of work done:-

$W = F \ • \ s \ cos(\theta)$

Answer 2

$\sf{Answer}$

Step by step explanation:-

Given :-

Mass of car :- 1800kg

Increase in velocity from 72km/hr - 90km/hr

To find :-

Work done by body

Formula implemented:-

W = ΔK.E = $\sf\dfrac{m}{2}$× $\sf{vf²-vi²}$

Solution:-

Work done by the body is equal to change in kinetic energy

So, velocity is given in km/hr

We have to convert into m/s

1km/hr = $\sf\dfrac{5}{18}$ m/s

72km/hr = $\sf\dfrac{5}{18}$×72 m/s = 20m/s

90km/hr = $\sf\dfrac{5}{18}$×90m/s= 25m/s

So, vi = 20m/s

vf = 25m/s

By using formula

W = ΔK.E = $\sf\dfrac{1}{2}$m $\sf{vf²-vi²}$

Plughing values :-

W = $\sf\dfrac{1}{2}$×1800×$\sf{25²-20²}$

W = $\sf{900}$×$\sf{625-400}$

W = $\sf{900× 225}$

W = $\sf{2,02,500J}$

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