A car is proceeding at a speed of 12.0 m/s when it collides with a stationary car in front. During the collision, the fist car moves a distance of 0.300 m as it comes to a stop. The driver is wearing her seat belt, so she remains in her seat during the collision. The driver’s mass is 56.0 kg. Neglect any friction between the driver and the seat.
1)How much force does the belt exert on her during the collision? (Express your answer to three significant figures.)
Well, well, well! Looks like we have a collision on our hands! Let's crunch those numbers, shall we?
To find out the force exerted by the seat belt on the driver, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration.
Since the car comes to a stop during the collision, we can calculate the acceleration using the kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given:
Initial velocity, u = 12.0 m/s (speed of the car before colliding)
Distance traveled, s = 0.300 m
Mass of the driver, m = 56.0 kg
First, we need to find the acceleration:
v^2 = u^2 + 2as
0^2 = 12.0^2 + 2a(0.300)
0 = 144 + 0.6a
Solving for 'a', we have:
0.6a = -144
a ≈ -240 m/s^2
Negative sign on the acceleration indicates the deceleration of the car.
Now, let's plug in the values into Newton's second law:
F = ma
F = 56.0 kg * -240 m/s^2
Calculating the force, we get:
F ≈ -13,440 N
So, it seems like the force exerted by the seat belt on the driver during the collision is approximately -13,440 Newtons. Remember, I neglected any friction between the driver and the seat, and please keep in mind that this answer is just an estimation based on the given data. Drive safely!
To find the amount of force exerted by the seat belt on the driver during the collision, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration.
First, we need to find the acceleration of the driver during the collision. Since the car comes to a stop and the driver remains in her seat, we can assume that her final velocity is 0 m/s. The initial velocity of the car is given as 12.0 m/s. We also have the displacement covered during the collision, which is 0.300 m.
Using the equation of motion (v² = u² + 2as), where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can rearrange it to solve for acceleration:
0 = (12.0 m/s)² + 2a(0.300 m)
0 = 144.0 m²/s² + 0.6a
Simplifying further:
0.6a = -144.0 m²/s²
a = -240.0 m²/s²
The negative sign indicates a deceleration since the car is coming to a stop.
Now that we have the acceleration, we can calculate the force using Newton's second law. The formula is F = m * a, where F is the force, m is the mass, and a is the acceleration.
F = (56.0 kg) * (-240.0 m²/s²)
F = -13,440 N
The negative sign indicates that the force is exerted in the opposite direction of the motion, in this case, decelerating the driver.
Therefore, the force exerted by the seat belt on the driver during the collision is approximately -13,440 N.