Question

to avoid an accident, a driver with a mass of 60kg applies his brakes and decelrates from 30m/s to 1m/s in 2 seconds. how much force was absorbed by his seat belt​

Answer 1

Answer:

$F=-870\ N$

Explanation:

Net Force

The net force applied to an object of mass m that moves at an acceleration a is given by:

F=m.a

The acceleration can be calculated by the equations of the constant acceleration motion:

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf is the final speed, vo is the initial speed, and t is the time.

Substituting the given values:

$\displaystyle a=\frac{1-30}{2}=\frac{-29}{2}=-14.5\ m/s^2$

The force absorbed by his seat belt:

$F=60\ kg\cdot (-14.5\ m/s^2)$

$\boxed{F=-870\ N}$

Answer 2

Force absorbed by the seatbelt is : 870 N

Given data :

mass of body = 60 kg

Initial velocity ( V₁ ) = 30 m/s

Final velocity ( V₂ ) = 1 m/s

Time = 2 secs

 

Determine the force absorbed by the seat belt

Force = Mass * acceleration  --- ( 1 )

where ;

acceleration = Δ v / t  ---- ( 2 )

Δ v = ( 30 - 1 ) = 29 m/s

t = 2 secs

Back to equation ( 2 )

acceleration = 29 / 2 = 14.5 m/s²

∴ Force exerted on seatbelt

F = m * a

  = 60 kg * 14.5 m/s²  =  870 N

Hence we can conclude that the force absorbed by his seatbelt is 870 N

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