An 800-N passenger in a car, presses against the car door with a 200 N force when the car makes a left turn at 13 m/s. The (faulty) door will pop open under a force of 800 N. Of the following, the least speed for which the man is thrown out of the car is:

Question

Quizplus · Accepted Answer

To solve this problem we need to calculate the centripetal force acting on the passenger when the car makes the left turn. The centripetal force is given by:

Fc = m*v^2/r

where m is the mass of the passenger, v is the speed of the car, and r is the radius of the turn. Since the turn is not banked, the radius of the turn is given by:

r = (mg - Fd)/Ff

where mg is the weight of the passenger, Fd is the force that the passenger exerts on the door, and Ff is the frictional force between the tires of the car and the road.

We are given that Fd = 200 N, and that the (faulty) door will pop open under a force of 800 N. Therefore, the maximum frictional force that the tires can provide is:

Ff = 800 N

We are also given that the weight of the passenger is:

mg = 800 N

Substituting these values into the equation for r, we get:

r = (800 N - 200 N)/800 N = 0.75 m

Now we can use the equation for Fc to find the minimum speed at which the passenger will be thrown out of the car. Since Fc must be greater than or equal to Fd for the passenger to stay in the car, we have:

Fc >= Fd

Substituting the values for Fd and r, we get:

m*v^2/r >= Fd

800 N*v^2/(0.75 m) >= 200 N

Solving for v, we get:

v >= sqrt(0.75 m*200 N/800 N) = 0.387 m/s

Therefore, the minimum speed at which the passenger will be thrown out of the car is:

Answer: D) 26 m/s

An 800-N Passenger in a Car, Presses Against the Car