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3 years ago

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A car weighing 19600N is moving with a speed of 30 m/sec on a level road. If it is brought to rest in a distance of 100 m. Find

the average distance friction force acting on it

1 answer:

goblinko [34]3 years ago

8 0

Answer:

F = -8820 N

Explanation:

Given that,

The weight of a car, W = 19600 N

Initial speed of the car, u = 30 m/s

It is brought to rest, final velocity, v = 0

Distance, d = 100 m

We need to find the average friction force acting on it.

Firstly we find the acceleration of the car using third equation of motion. Let it is a.

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(0)^2-(30)^2}{2\times 100}\\\\=-4.5\ m/s^2$

Average frictional force,

F = ma

m is mass, $m=\dfrac{W}{g}=\dfrac{19600\ N}{10\ m/s^2}=1960\ kg$

F = 1960 kg × -4.5 m/s²

= -8820 N

So, the average friction force acting on it is 8820 N.

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

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${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ <u>Using 1st equation of motion </u>

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

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$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

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☯ <u>Now, Finding the force exerted </u>

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$\dashrightarrow\:\: \sf{F = ma}$

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$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ <u>Hence</u>, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

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