In an emergency stop to avoid an accident,a shoulderstrap seat belt holds a 60-kg passenger firmly in place. If the car were initally traveling at 90 km/h and came to a stop in 5.5s along a straight, level road, what was the average force applied to the passenger by the seatbelt?
2.7×10²
To find the average force applied to the passenger by the seatbelt, we need to use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a).
First, let's convert the initial velocity of the car from km/h to m/s. Since 1 km/h is equal to 0.2778 m/s, we multiply 90 km/h by 0.2778 to get 25 m/s.
Next, let's calculate the acceleration of the car. We can use the formula:
a = (final velocity - initial velocity) / time
As the car comes to a stop, the final velocity is 0 m/s. So, the acceleration (a) is:
a = (0 - 25 m/s) / 5.5 s
a = -25 m/s / 5.5 s
a = -4.55 m/s²
Since the car is decelerating, the acceleration is negative.
Now, we can calculate the force applied to the passenger using Newton's second law:
F = m * a
Plugging in the values:
F = 60 kg * (-4.55 m/s²)
F ≈ -273 N
The negative sign indicates that the force is acting in the opposite direction to the motion of the car, which makes sense since it is stopping the passenger's forward motion.
Therefore, the average force applied to the passenger by the seatbelt is approximately 273 N.
First convert 90 km/h to ___meters per second.
The conversion factor is
(1000 m/km) * (1 hr/3600s)
Divide that speed in m/s by 5.5 s to get the deceleration rate in m/s^2.
Mass(kg) x acceleration (m/s^2)
= Force (Newtons)
If you want it in pounds, multiply Newtons by 0.225 lb/N