Seatbelts provide two main advantages in a car accident: (i) they keep you from being thrown from the car, and (ii) they reduce the force that acts on you during the collision to survivable levels. The second benefit can be illustrated by comparing the net force exerted on the driver of a car in a head-on collision with and without a seatbelt.
(a) A driver wearing a seatbelt decelerates at the same rate as the car itself. Since modern cars have a "crumple zone" built into the front of the car, the car will decelerate over a distance of roughly 1.0 m. Find the net force acting on a 62 kg driver who is decelerated from 19 m/s to rest in a distance of 1.0 m.
? kN
(b) A driver who does not wear a seatbelt continues to move forward with a speed of 19 m/s (due to inertia) until something solid is encountered. The driver now comes to rest in a much shorter distance -- perhaps only a centimeter. Find the net force acting on a 62 kg driver who is decelerated from 19 m/s to rest in 1.0 cm.
? kN
I am not sure how to set this problem up. Can someone describe the picture or suggest which equations to use?
a)
avg force *distance= 1/2 mass*initialvelocity^2
b) same formula, distance has changed.
a) We can calculate the net force acting on the driver using the work-energy theorem, which states that work done on an object is equal to its change in kinetic energy:
avg force * distance = 1/2 * mass * initial_velocity^2
We are given the initial velocity (19 m/s), mass of the driver (62 kg), and the distance over which the car decelerates (1 m). Plugging these values into the equation, we get:
avg force * 1.0 m = 1/2 * 62 kg * (19 m/s)^2
avg force = (1/2 * 62 kg * 361 m^2/s^2) / 1.0 m
avg force = 11221 kg m/s^2
To convert this force into kN, we divide the value by 1000:
avg force = 11.22 kN
Thus, the net force acting on the driver is 11.22 kN.
b) We use the same formula as in part (a), but now the distance over which the driver decelerates is only 1.0 cm or 0.01 m. Plugging this value into the equation, we get:
avg force * 0.01 m = 1/2 * 62 kg * (19 m/s)^2
avg force = (1/2 * 62 kg * 361 m^2/s^2) / 0.01 m
avg force = 1122100 kg m/s^2
To convert this force into kN, we divide the value by 1000:
avg force = 1122.1 kN
Thus, the net force acting on the driver is 1122.1 kN.
To solve this problem, we can use the equation for average force, which is:
Average Force = (1/2) * mass * (final velocity^2 - initial velocity^2) / distance
Let's calculate the net force acting on the driver in both cases:
(a) When the driver is wearing a seatbelt:
Given:
Mass of the driver (m) = 62 kg
Initial velocity (u) = 19 m/s
Final velocity (v) = 0 m/s (as they come to rest)
Deceleration distance (d) = 1.0 m
Using the equation, we get:
Average Force = (1/2) * 62 kg * ((0 m/s)^2 - (19 m/s)^2) / 1.0 m
= (1/2) * 62 kg * (-19 m/s)^2 / 1.0 m
Evaluating this expression, we find:
Average Force = 1109.5 N
Therefore, the net force acting on the driver, while wearing a seatbelt, is 1109.5 N.
(b) When the driver does not wear a seatbelt:
Given:
Mass of the driver (m) = 62 kg
Initial velocity (u) = 19 m/s
Final velocity (v) = 0 m/s (as they come to rest)
Deceleration distance (d) = 1.0 cm = 0.01 m (converted to meters)
Using the equation, we get:
Average Force = (1/2) * 62 kg * ((0 m/s)^2 - (19 m/s)^2) / 0.01 m
= (1/2) * 62 kg * (-19 m/s)^2 / 0.01 m
Evaluating this expression, we find:
Average Force = 122,600 N
Therefore, the net force acting on the driver, without wearing a seatbelt, is 122,600 N.
In summary:
(a) The net force acting on the driver, wearing a seatbelt, is 1109.5 N.
(b) The net force acting on the driver, without wearing a seatbelt, is 122,600 N.
To solve these problems, you can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the acceleration is the rate at which the car or the driver decelerates.
For part (a), since the driver wearing a seatbelt decelerates at the same rate as the car itself, the net force acting on the driver is the same as the net force acting on the car. The distance over which the car decelerates is given as 1.0 m. We can use the equation:
Force * distance = 0.5 * mass * velocity^2
Plugging in the known values:
Mass = 62 kg
Velocity = 19 m/s
Distance = 1.0 m
Rearranging the equation to solve for Force:
Force = (0.5 * mass * velocity^2) / distance
Substituting the values:
Force = (0.5 * 62 kg * (19 m/s)^2) / 1.0 m
Evaluating this expression will give you the net force acting on the driver in newtons (N).
For part (b), since the driver who does not wear a seatbelt comes to rest in a much shorter distance (1.0 cm), we can use the same equation as in part (a) with a different distance value. Calculate the force using the same formula, but substitute the new distance value of 1.0 cm (converted to meters) into the equation.
Finally, it is important to note that the force should be represented in kilonewtons (kN), so you will need to convert the values accordingly.