Question

A vehicle of mass 120 kg is moving with a uniform velocity of 108 km/hr. The force required to stop the vehicle in 10 sec is:

Answer 1

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HERE'S THE ANSWER...

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♠️ First we'll find out Acceleration , as

⏺️ Initial velocity ( u ) = 108 Km / h

=> 108 × 5 / 18

=> 30 m / s

⏺️ Final velocity ( v ) = 0

⏺️ Time ( t ) = 10 sec

✔️ We know

=> Acc. ( a ) = ( Final velocity - initial velocity ) / time

=> a = ( v - u ) / t

=> a = ( 0 - 30 ) / 10

=> a = - 30 / 10

=> $\boxed{ a\:= \: -3 \: m/s^2}$

♠️ Here Acceleration is negative be velocity is decreasing.

♠️ Now , We know force is product of mass and Acceleration , i.e

=> Force ( F ) = mass × acc ( a )

⏺️ Mass ( m ) = 120 Kg

⏺️ a = - 3 m/s^2

=> F = m × a

=> F = 120 × - 3

=> $\boxed{ Force\:=\:-360 \: Newton ( N )}$

♠️ Here force is negative because it is in opposite direction of motion.


HOPE HELPED.

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Answer 2

Here is your solution

$\huge\boxed{\red{\bold{Given:-}}}$

Initial velocity(U) of 108 km/hr.

to change into m/s

108×5/18
30m/s✔

final velocity (v)=0

Time (t)=10sec
$Acceleration = \frac{v - u}{t} \\ \\ a= \frac{0 - 30}{10} \\ \\ a = \frac{ - 30}{10} \\ \\ a = - 3m/ s {}^{2}$

Now

Force = mass × acceleration

Force=120×(-3)
$\huge\underline {\red{\bold{F= -360N}}}$

Note:-

➡Here force is negative because of opposite direction motion

➡The s.i unit of Force is newton

Hope it helps you

$\huge\boxed{\red{\bold{Thanks}}}$