Question

In an emergency stop, a shoulder- strap seat belt holds a 60 kg passenger firmly in place. The car was initially traveling at 30 m/s and came to a stop in 6.0 seconds along a straight, level road. What was the average force applied to the passenger by the seatbelt?

Answer 1

Mass (m) of the passenger = 60kg

Initial speed (u) of the car = 30km/h

                                      = (30)(5/18)m/s

                                       = 8.33 m/s

Final speed (v) of the car = 0

Time (t) taken to come to stop = 6.0 s

Now formula for the average force applied to the passenger bythe seatbelt is

                 F = m(v - u) / t

                    = (60kg)(0 - 8.33m/s) / (6.0s)

                    = -83.3N,   here negative sign indicates that the forceis applied in a direction opposite to that of motion.