A football is kicked at a speed of 18.0 m/s at an angle of 51.0 degrees to the horizontal. How much later does it hit the ground? What is the weight of a 68 kg astronaut (a) on Ear... A football is kicked at a speed of 18.0 m/s at an angle of 51.0 degrees to the horizontal. How much later does it hit the ground? What is the weight of a 68 kg astronaut (a) on Earth, (b) on the Moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²)? A 14.0 kg bucket is lowered vertically by a rope in which there is 63 N of tension at a given moment.

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Understand the Problem

The question involves physics problems related to projectile motion and weight in different gravitational fields. It requires calculations for the time of flight of a projectile, the weight of an astronaut on different celestial bodies, and the tension in a rope under certain conditions.

Answer

1. $666.4 \, \text{N}$ 2. $115.6 \, \text{N}$ 3. $251.6 \, \text{N}$ 4. $0 \, \text{N}$ 5. $137.2 \, \text{N}$
Answer for screen readers

  1. Weight on Earth: $666.4 , \text{N}$

  2. Weight on the Moon: $115.6 , \text{N}$

  3. Weight on Mars: $251.6 , \text{N}$

  4. Weight in outer space: $0 , \text{N}$

  5. Tension in the rope: $137.2 , \text{N}$

Steps to Solve

  1. Determine the weight of the astronaut on Earth

The weight of an object is calculated using the formula: $$ W = mg $$ where $W$ is weight, $m$ is mass, and $g$ is the acceleration due to gravity. For Earth, $g = 9.8 , \text{m/s}^2$.

Substituting the values: $$ W = 68 , \text{kg} \times 9.8 , \text{m/s}^2 $$

  1. Calculate the weight of the astronaut on the Moon

Using the same formula for weight, but with the Moon's gravitational acceleration of $g = 1.7 , \text{m/s}^2$: $$ W_{\text{Moon}} = 68 , \text{kg} \times 1.7 , \text{m/s}^2 $$

  1. Calculate the weight of the astronaut on Mars

Again using the weight equation with Mars' gravitational acceleration of $g = 3.7 , \text{m/s}^2$: $$ W_{\text{Mars}} = 68 , \text{kg} \times 3.7 , \text{m/s}^2 $$

  1. Identify the astronaut's weight in outer space

When an object is traveling in outer space at constant velocity, there is no net force acting on it (it is in a state of weightlessness), thus: $$ W_{\text{space}} = 0 , \text{N} $$

  1. Review the total tension in the rope for the bucket

The equation for tension in the rope, when the bucket is lowered at a constant velocity, is equal to the weight of the bucket: $$ T = W_{\text{bucket}} = mg $$ For the 14.0 kg bucket: $$ T = 14.0 , \text{kg} \times 9.8 , \text{m/s}^2 $$

  1. Weight on Earth: $666.4 , \text{N}$

  2. Weight on the Moon: $115.6 , \text{N}$

  3. Weight on Mars: $251.6 , \text{N}$

  4. Weight in outer space: $0 , \text{N}$

  5. Tension in the rope: $137.2 , \text{N}$

More Information

The calculations illustrate how weight varies based on the local gravitational field strength. The astronaut's weight is much less on the Moon due to lower gravity and weightless in space due to the lack of net force. The bucket shows that tension in a rope under constant velocity matches the weight of the object.

Tips

  • Forgetting to use the correct gravitational acceleration for each location can lead to incorrect weight calculations.
  • Confusing weight with mass; weight changes with gravity while mass remains constant.
  • Misunderstanding the concept of weightlessness in outer space, where objects do not experience weight due to free-fall.

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