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Math  /  Algebra

Questionthe magnitude of its acceleration. A rocket sled accelerates at a rate of 49.0 m/s249.0 \mathrm{~m} / \mathrm{s}^{2}. 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. Repeat the previous problem for a situation in which the rocket sled decelerates at a rate of 201 m/s2201 \mathrm{~m} / \mathrm{s}^{2}. In this problem, the forces are exerted by the seat and the seat belt. A body of mass 2.00 kg is pushed straight upward by a 25.0 N vertical force. What is its acceleration?

Studdy Solution

STEP 1

1. The rocket sled accelerates at 49.0 m/s2 49.0 \, \mathrm{m/s}^2 .
2. The passenger has a mass of 75.0 kg 75.0 \, \mathrm{kg} .
3. The gravitational acceleration is 9.8 m/s2 9.8 \, \mathrm{m/s}^2 .
4. For the second part, the rocket sled decelerates at 201 m/s2 201 \, \mathrm{m/s}^2 .
5. A body of mass 2.00 kg 2.00 \, \mathrm{kg} is pushed upward by a 25.0 N 25.0 \, \mathrm{N} force.

STEP 2

1. Calculate the horizontal component of the force and compare it with the weight.
2. Calculate the direction and magnitude of the total force.
3. Repeat the calculations for deceleration.
4. Calculate the acceleration of the body pushed upward.

STEP 3

Calculate the horizontal component of the force exerted by the seat on the passenger.
The force exerted by the seat is given by Newton's second law: F=ma F = ma where m=75.0 kg m = 75.0 \, \mathrm{kg} and a=49.0 m/s2 a = 49.0 \, \mathrm{m/s}^2 .
F=75.0×49.0=3675 N F = 75.0 \times 49.0 = 3675 \, \mathrm{N}

STEP 4

Compare this force with the weight of the passenger.
The weight W W of the passenger is: W=mg=75.0×9.8=735 N W = mg = 75.0 \times 9.8 = 735 \, \mathrm{N}
The ratio of the force to the weight is: Ratio=FW=3675735≈5.0 \text{Ratio} = \frac{F}{W} = \frac{3675}{735} \approx 5.0

STEP 5

Calculate the direction and magnitude of the total force exerted by the seat.
Since the force is horizontal, the direction is along the horizontal axis. The magnitude is already calculated as 3675 N 3675 \, \mathrm{N} .

STEP 6

Repeat the calculations for deceleration at 201 m/s2 201 \, \mathrm{m/s}^2 .
The force exerted by the seat and seat belt is: Fdeceleration=ma=75.0×201=15075 N F_{\text{deceleration}} = ma = 75.0 \times 201 = 15075 \, \mathrm{N}

STEP 7

Compare this force with the weight of the passenger.
The ratio of the force to the weight is: Ratio=FdecelerationW=15075735≈20.5 \text{Ratio} = \frac{F_{\text{deceleration}}}{W} = \frac{15075}{735} \approx 20.5

STEP 8

Calculate the acceleration of a body of mass 2.00 kg 2.00 \, \mathrm{kg} pushed upward by a 25.0 N 25.0 \, \mathrm{N} force.
The net force acting on the body is the applied force minus the weight: Fnet=25.0−(2.00×9.8)=25.0−19.6=5.4 N F_{\text{net}} = 25.0 - (2.00 \times 9.8) = 25.0 - 19.6 = 5.4 \, \mathrm{N}
The acceleration a a is given by: a=Fnetm=5.42.00=2.7 m/s2 a = \frac{F_{\text{net}}}{m} = \frac{5.4}{2.00} = 2.7 \, \mathrm{m/s}^2

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