Question

A 50 kilogram woman wearing a seatbelt is traveling in a car that is moving with a velocity of +10 meters per second. In an emergency, the car is brought to a stop in 0.50 seconds. What force does the seatbelt exert on the woman so that she remains in her seat?

A. -1.0 x 10^3 N
B. -5.0 x 10^2 N
C. -5.0 x 10^1 N
D. -2.5 x 10^1 N

Answer 1

Answer:

the answer is A.) -1 * 10^3[N]

Explanation:

The solution consists of two steps, the first step is using the following kinematic equation:

$v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]$

The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].

Replacing:

$0=10+a*(0.5)\\a=-20[m/s^2]$

Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.

In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.

$F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]$

The minus sign means that the force is acting in the opposite direction of the movement.