Question

A car going 15 m/s is brought to rest in a distance of 2.0 m as it strikes a pile of dirt. How large an average force is exerted by seatbelts on a 90 kg passenger as the car is stopped?

Answer 1

Answer:

Given data

Vi=15 m/s

Vf=0 m/s

S= 2m

m=90 kg

Required

Force of seatbelt on passenger =?

Explanation:

If car has velocity of 15m/s and cover 2m distance before to rest. So the time will be

t=2/15 => t=0.13s

to find the acceleration of the car

a=vf-vi/t => a= -115. 4m/s^2

Here minus sign shows deceleration

So, speed for the car and man will be same and acceleration also

F=ma

F=90*115.4

F= 10.4 kN

Answer 2

Concept :

The push or pull that changes the speed of a mass-containing thing. Force is an outside specialist that has the ability to modify a body's resting or moving position. It has a course and a greatness. There are a few cases of powers in day to day existence, including: haul and power (for example the heaviness of something) the power a bat applies to a ball. the tension that a hair brush applies to hair while brushing it. the tension your foot applies to the pedal while you're riding a bicycle.

Given:

A car going $15 m/s$ is brought to rest in a distance of $2.0 m$ as it strikes a pile of dirt.

To Find:

The average force.

Solution:

According to the problem,

$t=2/15 = > t=0.13s$

to find the acceleration of the car

$a=vf-vi/t = > a= -115. 4m/s^2$

Here minus sign shows deceleration

So, speed for the car and man will be same and acceleration also

$F=ma\\F=90*115.4\\F= 10.4kN$

Hence the force will be $10.4 kN$

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