Question

In an emergency stop to avoid an accident, a shoulder-strap seatbelt holds a 62 kg passenger in place. If the car was initially traveling at 88 km/h and came to a stop in 6.0 s along a straight, level road, what was the magnitude of the average force applied to the passenger by the seatbelt?

a) 620 N
b) 720 N
c) 820 N
d) 920 N

Answer 1

The magnitude of the average force applied to the passenger by the seatbelt is approximately 249.95 N.

Step-by-step explanation:

To calculate the force applied to the passenger by the seatbelt, we can use the equation for average force, which is given by F = ma. Given that the mass of the passenger is 62 kg and the car came to a stop in 6.0 s, we can calculate the average force applied to the passenger using the equation: F = mΔv/Δt, where Δv is the change in velocity and Δt is the change in time. First, we need to calculate Δv. The initial velocity is given as 88 km/h, which we need to convert to m/s by multiplying it by 1000/3600. So, the initial velocity is 88 km/h × (1000 m/3600 s) = 24.44 m/s. The final velocity is 0 m/s since the car came to a stop. Therefore, Δv = 0 - 24.44 m/s = -24.44 m/s. Now we can substitute the values into the equation to calculate the average force: F = (62 kg)(-24.44 m/s) / 6.0 s = -249.95 N. Since the question asks for the magnitude of the average force, we take the absolute value of the force, which is 249.95 N. Therefore, the magnitude of the average force applied to the passenger by the seatbelt is approximately 249.95 N.