How much later does it hit the ground? What is the weight of a 68 kg astronaut on Earth, on the Moon, on Mars, and in outer space traveling with constant velocity? A 14.0 kg bucket... How much later does it hit the ground? What is the weight of a 68 kg astronaut on Earth, on the Moon, on Mars, and in outer space traveling with constant velocity? A 14.0 kg bucket is lowered vertically by a rope with 163 N of tension. What is the acceleration of the bucket? Is it up or down?
Understand the Problem
The question is asking for the calculations of weight on different celestial bodies, and the acceleration of a bucket based on tension in a rope. We'll solve these physics problems using the respective gravitational values for Earth, Moon, and Mars as well as applying Newton’s second law to find the acceleration of the bucket.
Answer
Weight on Earth: $W_{Earth} = m \cdot 9.81$, weight on Moon: $W_{Moon} = m \cdot 1.62$, weight on Mars: $W_{Mars} = m \cdot 3.71$, acceleration of bucket: $a = \frac{F}{m}$.
Answer for screen readers
The weight on each celestial body can be expressed as:
- $W_{Earth} = m \cdot 9.81$
- $W_{Moon} = m \cdot 1.62$
- $W_{Mars} = m \cdot 3.71$
The acceleration of the bucket can be calculated given a specific tension value using:
$$ a = \frac{F}{m} $$
Steps to Solve
- Identify the gravitational values
We need to gather the gravitational acceleration values for Earth, Moon, and Mars.
These are approximately:
- Earth: $g_{Earth} = 9.81 , \text{m/s}^2$
- Moon: $g_{Moon} = 1.62 , \text{m/s}^2$
- Mars: $g_{Mars} = 3.71 , \text{m/s}^2$
- Calculate weight on each celestial body
Weight can be calculated using the formula:
$$ W = m \cdot g $$
where ( W ) is weight, ( m ) is mass, and ( g ) is gravitational acceleration.
We will need the mass in kilograms to solve.
- Example calculation for Earth
For a given mass ( m ), the weight on Earth would be:
$$ W_{Earth} = m \cdot g_{Earth} = m \cdot 9.81 $$
- Repeat for Moon and Mars
We use the same formula for Moon and Mars:
$$ W_{Moon} = m \cdot g_{Moon} = m \cdot 1.62 $$
$$ W_{Mars} = m \cdot g_{Mars} = m \cdot 3.71 $$
- Determine acceleration of the bucket
Using Newton’s second law, the force exerted by the tension in the rope can be calculated as:
$$ F = m \cdot a $$
where ( a ) is the acceleration, and ( F ) can be determined from the tension in the rope.
To find the acceleration, rearrange the formula:
$$ a = \frac{F}{m} $$
The weight on each celestial body can be expressed as:
- $W_{Earth} = m \cdot 9.81$
- $W_{Moon} = m \cdot 1.62$
- $W_{Mars} = m \cdot 3.71$
The acceleration of the bucket can be calculated given a specific tension value using:
$$ a = \frac{F}{m} $$
More Information
The weights calculated are dependent on mass ( m ). If you know the mass, you can simply plug that value into the formulas to get your answers. It's interesting to note how much lighter objects are on the Moon compared to Earth!
Tips
- Forgetting to convert units if mass is not in kilograms, leading to incorrect calculations.
- Misapplying the formula for weight or acceleration by mixing up the values for mass and gravitational acceleration.
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