Final answer:
To find the force of the seat belt on the passenger, deceleration is first determined using the kinematic equation, and then Newton's second law is applied to calculate force. With a deceleration of -8.60 m/s² and a mass of 80.0 kg, the calculated force is 688 N, So, the best answer is e, 688 N.
Step-by-step explanation:
To calculate the force of the seat belt on the passenger, we need to use the concepts of deceleration and Newton's second law of motion. We start by converting the SUV's initial speed from kilometers per hour to meters per second:
Initial speed (vi) = 1.00 × 102 km/h
= 100 km/h
= (100 × 1000 m) / (3600 s)
= 27.78 m/s
The SUV comes to a complete stop, which means the final speed (vf) is 0 m/s. Using the equation for acceleration:
a = (vf - vi) / t
However, we do not have the time (t). Instead, we can use the kinematic equation that relates initial speed, final speed, acceleration, and distance:
vf2 = vi2 + 2ad
0 = (27.78 m/s)2 + 2a(45.0 m)
Solving for acceleration (a), we find:
a = -(27.78 m/s)2 / (2 × 45.0 m)
= -8.60 m/s2
The negative sign indicates deceleration. Finally, using Newton's second law, F = ma, where m is the mass of the passenger (80.0 kg) and a is the deceleration:
F = 80.0 kg × 8.60 m/s2
= 688 N
However, this force is the net force required to stop the passenger.
The actual force exerted by the seatbelt would be equal and opposite to this net force, so the magnitude is the same: 688 N.
So, the best answer is e, 688 N.
Q: An 80.0-kg passenger in an SUV traveling at 1.00×102 km/h is wearing a seat belt. The driver slams on the brakes and the SUV stops in 45.0 m. Find the force of the seat belt on the passenger.
a) 3,125 N
b) 4,444 N
c) 5,333 N
d) 6,250 N
e) 688 N.