Question

n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration on Zuton is 3.0 m/s 2 (down). At the moment of liftoff the acceleration of the space ship is 0.50 m/s 2 (up). What is the magnitude of the force of the space ship on the astronaut

Answer 1

Answer: 0.29 kN

Explanation:

We have the following data:

$W_{E}=800 N$ is the weight of the astronaut on Earth

$g_{E}=9.8 m/s^{2}$ is the free fall acceleration due gravity on Earth (directed downwards)

$g_{Z}=3 m/s^{2}$ is the free fall acceleration due gravity on Zuton (directed downwards)

$a=0.5 m/s^{2}$ is the acceleration of the spaceship at litoff (directed upwards)

We have to find the magnitude of the force $F$ the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

$W_{E}=mg_{E}$ (1)

Where $m$ is the mass of the atronaut.

Isolating $m$:

$m=\frac{W_{E}}{g_{E}}$ (2)

$m=\frac{800 N}{9.8 m/s^{2}}$ (3)

$m=81.63 kg$ (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

$W_{Z}=mg_{Z}$ (5)

$W_{Z}=(81.63 kg)(3 m/s^{2})$ (6)

$W_{Z}=244.89 N$ (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

$F-W_{Z}=m.a$ (8)

Isolating $F$:

$F=m.a+W_{Z}$ (9)

$F=(81.63 kg)(0.5 m/s^{2})+244.89 N$ (10)

$F=285.7 N \frac{1 kN}{1000 N}=0.285 kN$ (11)

Finally:

$F=0.285 kN \approx 0.29 kN$