Question

A car leaves the road traveling at 110 km/h and hits a tree, coming to a stop in 0.14 s. What average force does a seatbelt exert on a 60-kg passenger during this collision? Wolfson, richard. Essential university physics, volume 1 (page 67). Pearson education. Kindle edition.

Answer 1

We use the Newtons law,

$F=m a$                        (A)

Equation of motion,

$v = u + at$                   (B)

Here, v is final velocity , u is initial velocity and a is acceleration.

Given  $u = 110 \ km/h = 110 \frac{10^3 m}{60 \times 60 \ s} =30.55 \ m/s, t = 0.14 s$ .

As car coming to stop, so final velocity becomes zero i.e  $v = 0$.

Substituting these values in equation (B), we get

$0 = 33 . 55 \ m/s + a (0.14 \ s ) \\\\ a = \frac{ (-33.50 m/s )}{0.14 s } =- 239 .64 \ m/s^2$

Here, negative sign means, deceleration to rest.

As given the mass of passenger, m = 60 kg.

Therefore, from equation (A),

$F = (60 kg) (-239 .64 \ m/s^2) = -14378.57 N$

Thus , the force seat belt exert on the passenger during the collision is $-14378.57 \ N$