Question

Rhonda, who has a mass of 60.0kg, is riding at 25.0 m/s in her sports car when she must suddenly slam on the brakes to avoid hitting a dog crossingthe road. She is wearing her seatbelt, which brings her body to a stop in0.400 s. a) What average force did the seatbelt exert on her? b) If she had not been wearing her seatbelt, and the windshield had stopped her head in 1.0 x 10⁻³s, what average force would the windshield have exerted on her?c) How many times greater is the stopping force of the windshield than theseatbelt?

Answer 1

ANSWERS

(a) 3750 N

(b) 1.5 · 10⁶ N

(c) 400 times

EXPLANATION

(a) The force the seatbelt exerts on her is given by Newton's second law,

$F=m\cdot a$

Rhonda's average acceleration is her speed divided by the brake time,

$F=m\cdot\frac{\Delta v}{\Delta t}$

For this problem,

• m = 60.0 kg

,

• Δv = 25.0 m/s

,

• Δt = 0.400 s

$F=60.0\operatorname{kg}\cdot\frac{25.0m/s}{0.400s}=3750N$

The average force exerted by the seatbelt on Rhonda is 3750 N.

(b) The same principle applies in this case. This time, Δt = 1.0 · 10⁻³ s, instead of 0.4s as before,

$F=60.0\operatorname{kg}\cdot\frac{25.0m/s}{0.001s}=1500000N=1.5\times10^6N$

The force the windshield exerted on her was 1.5 · 10⁶ N.

(c) To find how many times greater the stopping force of the windshield is than the seatbelt, we have to divide the force found in part b by the force found in part a,

$\frac{1.5\times10^6N}{3750N}=400$

The stopping force of the windshield is 400 times greater than the seatbelt force.