Question

a car of mass 500 kg moving with the 72 kilometre per hour how much force is required to stop the car within 100 metre​

Answer 1

Mass of the car (m) = 500 kg

Initial velocity of the car (u) = 72 km/h = 20 m/s

★  1 km/h = ⁵/₁₈ m/s

As it stops finally , so

Final velocity of the car (v) = 0 km/h = 0 m/s

Distance moved by the car (s) = 100 m

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Force (F) : It is defined as mass times the acceleration . SI unit is Newton .

➳ F = m × a

3rd equation of motion :

➳ v² - u² = 2as

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➠ v² - u² = 2as

➠ (0)² - (20)² = 2a(100)

➠ 0 - 400 = 200a

➠ 200a = -400

➠ a = - 2 m/s²

Note : - ve sign of acceleration denotes retardation

$:\implies \sf F=ma$

$:\implies \sf F=(500)(-2)$

$:\implies \sf{\bf{\pink{F=-1000\ N}}}\ \; \bigstar$

Note : - ve sign of force denotes retarding force

Magnitude of the force = 1000 N

Answer 2

Answer:

• Mass of Car ( m ) = 500 Kg
• Initial Speed ( u ) = 72 km/h
• Final Speed ( v ) = 0 [ As Car need to Stop ]
• Force required to stop car within 100 metres

$\underline{\bigstar\:\textsf{Using Third Equation of Motion :}}$

$:\implies\sf v^2-u^2=2as\\\\\\:\implies\sf (0)^2-(72\:km/h)^2=2 \times a \times 100\:m\\\\\\:\implies\sf -\:\bigg(72 \times \dfrac{5}{18}\:m/s\bigg)^2=2 \times a \times 100\:m\\\\\\:\implies\sf -\:(4 \times 5\:m/s)^2 = 2 \times a \times 100\:m\\\\\\:\implies\sf - \:(20\:m/s)^2 = 2 \times a \times 100\:m\\\\\\:\implies\sf - \:400 \:m^2/s^2 = 200 \:m \times a\\\\\\:\implies\sf \dfrac{- \:400 \:m^2/s^2}{200 \:m} = a\\\\\\:\implies\sf a = - \:2 \:m/s^2$

$\rule{180}{1.5}$

$\underline{\bigstar\:\textsf{Using formula for Force :}}$

$:\implies\sf Force = Mass \times Acceleration\\\\\\:\implies\sf Force=500\:kg \times -\:2\:m/s^2\\\\\\:\implies\sf Force=-\:1000\:kg\:m/s^2\\\\\\:\implies\underline{\boxed{\sf Force=-\:1000\:N}}$

⠀

$\therefore\:\underline{\textsf{It'll take \textbf{- 1000 N} to stop car within 100 metres}}.$