5. Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid hitting a deer crossing the road. She strikes the air bag, that brings her body to a stop in 0.500 s. What average force does the seat belt exert on her?
To find the average force exerted by the seat belt on Jennifer, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = m * a.
First, let's find the acceleration of Jennifer's body. Since she comes to a stop, we can assume that her final velocity is zero (v = 0), and her initial velocity is given as 35.0 m/s (u = 35.0 m/s). The time taken to stop is given as 0.500 s (t = 0.500 s).
To find the acceleration (a), we can use the equation: a = (v - u) / t
Substituting the given values, we get: a = (0 - 35.0) m/s / 0.500 s
Simplifying, we have: a = -70 m/s^2
Now that we have the acceleration, we can calculate the force exerted by the seat belt: F = m * a
Substituting the given mass of Jennifer (m = 50.0 kg) and the calculated acceleration (a = -70 m/s^2), we get: F = 50.0 kg * (-70 m/s^2)
Solving this equation, we find: F = -3500 N
The average force exerted by the seat belt on Jennifer is -3500 N (negative because it opposes her motion).