Question

1. In the car that decelerates from 30 kph to rest over 15 cm are two 70 kg passengers. One wears a seat belt and decelerates over 30 cm (with respect to the ground) (two significant figures). The other is not wearing a sea tbelt and decelerates over 5 cm* (with respect to the ground). What are the magnitudes of the average external forces acting on the two passengers during their de celerations? (Careful with significant figures: do both answers have the same precision?)

a) With a seat belt? _____ kN.

b) Without a seat belt? _____ kN.

Answer 1

As per the given question,

The initial speed of the car $u=30kph= 30\times 1000/3600=8.33m/sec$

Final velocity of the ca (v) = 0m/sec

Distance covered by the car during retardation(d)=15cm= 0.15m

so retardation $(a)$ , and weight of the man, who is inside the car is m1=m2=70kg and

displacement of person 1 who has weared the seat belt(d1)=30cm

and displacement of person 2 who has not weared the seat belt (d2)=5 cm

Now, $v^2=u^2-2ax$

$\Rightarrow a=\dfrac{8.33^2}{2\times0.15}=231.29\sim 231.30m/sec^2$

a) Pseudo force on the person 1 = m1a = $71\times231.30 N,$

person 1, traveling 30 cm distance , so retardation acceleration on the person who has weared the seat belt = $a_1=\dfrac{v^2}{2\times d_1}=\dfrac{8.33\times 8.33}{2\times 0.3}=115.648\sim115.65 m/sec^2$

so, net external acceleration on the man $=ma-ma_1=70(231.30-115.65)=8095.5N \simeq 8.10KN$

b) Force applied by the person who has not weared the seat belt $a_2=\dfrac{v^2}{2d_2}=\dfrac{8.33\times 8.33}{2\times0.05}=693.889\simeq693.89 m/sec^2$

Now, Net external force force applied by the person = $ma+ma_2=70\times231.30 +70\times693.89 =64763.3N\simeq64.78 KN$