in an emergency stop to avoid an accident, a shoulder-strap seat belt holds a 60kg passengr firmly in place. if the car was initially travelling at 90km/h and came to a stop in 4.5s along a straight, level road, what was the average force applied to the passenger by the seat belt? please help
Find the acceleration (deltaVeloicty/time)
Force=m*acceleration
so i converted my 90km/h to 25m/s, then found acceleration which was -5.56m/s^2.. i put that into F=m*a which was F=(60kg)(-5.56) F=-3336N or in significant digits -3000N.. is that correct?
55555577657374383284894270000000 N
To find the average force applied to the passenger by the seat belt, we can use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = ma).
First, we need to find the acceleration of the car. We can use the formula for acceleration: acceleration (a) = change in velocity / time taken (Δv / t).
Given:
Initial velocity (u) = 90 km/h = 90,000 m/3600 s = 25 m/s
Final velocity (v) = 0 m/s (since the car comes to a stop)
Time taken (t) = 4.5 s
Change in velocity (Δv) = v - u = 0 - 25 = -25 m/s (negative sign indicates deceleration)
Now, we can substitute these values into the acceleration formula to calculate the acceleration:
a = Δv / t = -25 m/s / 4.5 s = -5.56 m/s^2
Since the passenger is held firmly in place, the seat belt exerts a force on them equal to mass multiplied by the acceleration. We are given the mass of the passenger as 60 kg.
F = ma = 60 kg * -5.56 m/s^2 = -333.33 N
The negative sign indicates that the force is acting in the opposite direction of motion, which is the direction of deceleration in this case. To find the force without the negative sign, we take the magnitude:
Magnitude of force = |-333.33 N| = 333.33 N
Therefore, the average force applied to the passenger by the seat belt is approximately 333.33 Newtons.