Question

Repeat the previous problem for a situation in which the rocket sled accelerates opposite to the motion at a rate of 201 m/s². In this problem, the forces are exerted by the seat and the seat belt.

Reference Problem:
A rocket sled accelerates at a rate of 49.0 m/s². Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

a) What is the horizontal component of the force exerted by the seat against the passenger's body?
a) 15,075 N
b) 15,075 N in the opposite direction
c) 15,075 N in the same direction
d) 15,750 N

b) What is the direction and magnitude of the total force exerted by the seat against the passenger?
a) 15,075 N downward
b) 15,075 N upward
c) 15,075 N in the direction of motion
d) 15,750 N downward

Answer 1

The horizontal component and the total force exerted by the seat on the passenger are both 15,075 N, directed opposite to the deceleration and in the original direction of motion.

Step-by-step explanation:

For the scenario where a rocket sled decelerates at a rate of 201 m/s², and the passenger has a mass of 75.0 kg, we need to calculate the horizontal component of the force the seat exerts on the passenger and the direction and magnitude of the total force exerted by the seat.

To find the horizontal component of the force (Fh), we use Newton's second law: Fh = ma, where m is the mass of the passenger and a is the acceleration (or deceleration in this case). Substituting the given values, we get: Fh = (75.0 kg) (201 m/s²), which equals 15,075 N. The force exerted by the seat will be in the opposite direction to the deceleration, which is also the original direction of motion.

The magnitude of the total force (Ft) exerted by the seat is the same as the horizontal component because there is no vertical acceleration, thus Ft = Fh which is 15,075 N.

The correct answers are:

• a) The horizontal component of the force exerted by the seat against the passenger's body is 15,075 N in the opposite direction to the deceleration.
• b) The direction and magnitude of the total force exerted by the seat against the passenger is 15,075 N in the direction of the initial motion.