Question

a car of mass 150 kg moving at 20m/s is brought to 10m/s in 5sec . Find the force acting on the car?​

Answer 1

Answer:

Given

m= 150 kg

u= 20 m/sec

v= 10m /sec

t= 5 sec

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TO FIND

force=?

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F= m.a

F= m.aa= v-u/t

a= 10-20/5

a= -10/5

a= -2m/sec²

F= 150 . -2 = - 300 Newton

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Answer 2

Answer:

Question

a car of mass 150 kg moving at 20m/s is brought to 10m/s in 5sec . Find the force acting on the car?

Given

• Mass = 150 kg
• inital Velocity = 20 m/sec
• final Velocity =10 m/sec
• time = 5 sec

To find

Find the force acting on the car.

Solution

According to Newton's second law motion:-

${ \underline{ \boxed { \sf \color {aqua}{f = m.a}}}}$

where

• f denotes force
• m denotes mass
• a denotes accleration

But we need to find acceleration in order to find force-

according to first law of motion -:

${ \underline{ \boxed{ \sf \color{blue} v = u + at }}}$

where

• v denotes final velocity
• u denotes inital Velocity
• a denotes accleration
• t denotes time

subsitute the values in first law of motion

$\sf{ \implies \: 10 = 20 + a \times 5} \\ \\ \sf{ \implies \: - 10 = 5.a } \\ \\ \sf{ \implies \: a = \frac{ - 10}{5} } \\ \\ \sf{ \color{magenta}\implies \: a = - 2}$

* note - here (-) sign indicate the retardation of the car ..

Now substitute the value of accleration in f= m.a

$\sf{ \implies \: f = 150 \times ( - 2)} \\ \\ \sf{\implies \color{gold} \: f = ( - )300N}$

*note - (-) sign indicates the retarding force acting on the moving car to bring it to rest.

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More to know !!

• v= u+at
• s= ut+1/2 at²
• v²= u²+2as

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