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At the surface of Jupiter's moon Io, the acceleration due to gravity is \(g=1.81 \mathrm{~m} / \mathrm{s}^{2}\). A watermelon weighs \(44.0 \mathrm{~N}\) at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? (b) What would be its mass and weight on the surface of Io?

Short Answer

Expert verified
The mass of the watermelon on Earth would be approximately 4.48 kg. The same watermelon would still have a mass of 4.48 kg on Io, but its weight would decrease to 8.11 N due to the lower gravity.

Step by step solution

01

Calculate the mass on Earth

We know that weight = mass * gravity. Rearranging this formula we can find the mass on Earth. \(Mass = Weight/g\). Given the weight on Earth is 44.0 N and g on Earth is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\). We substitute these values into the formula.
02

Calculate the weight on Io

Now that we know the mass of the object, we can find the weight of the object on Io using the formula weight = mass * gravity. Given that g on Io is \(1.81 \mathrm{~m} / \mathrm{s}^{2}\), we substitute these values in to determine the weight on Io.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration Due to Gravity
Understanding the concept of acceleration due to gravity is foundational in gravitational physics. This acceleration is a measure of how quickly an object picks up speed as it falls toward the surface of a celestial body, due to the body’s gravitational pull. On Earth, this acceleration is approximately 9.81 meters per second squared (\(9.81 \text{ m/s}^{2}\)), and it is denoted by the symbol 'g'.

Importantly, this value is not constant across all celestial bodies—it varies based on the body's mass and radius. For instance, on Jupiter's moon Io, 'g' is 1.81 meters per second squared (\(1.81 \text{ m/s}^{2}\)). This means that an object on Io would accelerate at a slower rate compared to Earth because Io's gravitational pull is weaker.
Weight and Mass Relationship
The relationship between weight and mass is one of the pivotal ideas in physics. Mass is a measure of the amount of matter in an object, typically measured in kilograms. It remains constant regardless of location. Weight, on the other hand, is the force exerted on an object due to gravity and is measured in newtons. The formula that connects these quantities is simple: Weight = Mass × Gravity (\(W = m \times g\)).

In our exercise, to find the mass of a 44.0 newton watermelon on Earth, we rearrange the formula to Mass = Weight ÷ Gravity (\(m = W/g\)). This formula is key for converting weight (a force) to mass (a scalar quantity), provided we know the value of 'g'. The watermelon’s mass doesn’t change when it is transported to Io; however, its weight does because Io’s 'g' is less than Earth's.
Gravitational Acceleration on Different Celestial Bodies
Gravitational acceleration on different celestial bodies is an enthralling topic because it so clearly demonstrates the diversity of our solar system. Each celestial body has its own value of 'g', influenced by its mass and the distance from its center to the surface. Larger planets like Jupiter have a higher 'g' compared to smaller planets like Mars or moons such as Io. This variance has practical implications, such as how much a spacecraft needs to accelerate to escape a planet’s gravity or how much an astronaut can lift on a different planet or moon.

For example, the watermelon that weighs 44.0 N on Earth would have a different weight on Io. After finding the mass, as we did in the exercise, we use that same mass to calculate the new weight on Io (\(W = m \times g\text{ on Io}\)). Weight on Io is less because of its lower 'g', not because the watermelon's mass has changed. This illustrates how gravity not only shapes the weight of objects but also profoundly affects the surface conditions on various celestial bodies.

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Most popular questions from this chapter

Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block \(A\) has mass \(15.0 \mathrm{~kg},\) and block \(B\) has mass \(m\). A constant horizontal force \(F=60.0 \mathrm{~N}\) is applied to block \(A\) (Fig. \(\mathbf{P 4 . 3 8}\) ). In the first 5.00 s after the force is applied, block \(A\) moves \(18.0 \mathrm{~m}\) to the right. (a) While the blocks are moving, what is the tension \(T\) in the rope that connects the two blocks? (b) What is the mass of block \(B ?\)

A .22 caliber rifle bullet traveling at \(350 \mathrm{~m} / \mathrm{s}\) strikes a large tree and penetrates it to a depth of \(0.130 \mathrm{~m}\). The mass of the bullet is \(1.80 \mathrm{~g}\). Assume a constant retarding force. (a) How much time is required for the bullet to stop? (b) What force, in newtons, does the tree exert on the bullet?

A chair of mass \(12.0 \mathrm{~kg}\) is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F=40.0 \mathrm{~N}\) that is directed at an angle of \(37.0^{\circ}\) below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude \(15 \mathrm{~m} / \mathrm{s}^{2} .\) How much horizontal force must a \(55 \mathrm{~kg}\) sprinter exert on the starting blocks to produce this acceleration? Which object exerts the force that propels the sprinter: the blocks or the sprinter herself?

You walk into an elevator, step onto a scale, and push the "up" button. You recall that your normal weight is \(625 \mathrm{~N}\). Draw a free-body diagram. (a) When the elevator has an upward acceleration of magnitude \(2.50 \mathrm{~m} / \mathrm{s}^{2},\) what does the scale read? (b) If you hold a \(3.85 \mathrm{~kg}\) package by a light vertical string, what will be the tension in this string when the elevator accelerates as in part (a)?

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