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Chapter 4: Problem 15

IE \(\bullet\) A \(6.0-\mathrm{kg}\) object is brought to the Moon, where the acceleration due to gravity is only one-sixth of that on the Earth. (a) The mass of the object on the Moon is (1) zero, \((2) 1.0 \mathrm{~kg},(3) 6.0 \mathrm{~kg}\) (4) 36 kg. Why? (b) What is the weight of the obiect on the Moon?

Short Answer

Expert verified
(a) The mass on the Moon is 6.0 kg; (b) The weight on the Moon is approximately 9.78 N.

Step by step solution

01

Understanding Mass on the Moon

Mass is an intrinsic property of an object and does not change when the object is moved from one place to another. Thus, the mass of an object remains the same whether it is on the Earth, the Moon, or anywhere else in the universe.
02

Determining Mass Based on the Given Options

We are given several options for the mass of the object on the Moon: (1) zero, (2) 1.0 kg, (3) 6.0 kg, (4) 36 kg. Since mass does not change based on location, the mass of the object on the Moon is the same as on Earth. Therefore, the correct answer is option (3) 6.0 kg.
03

Understanding Weight on the Moon

Weight is the force exerted on an object due to gravity and is calculated using the formula: \[ \text{Weight} = \text{mass} \times \text{acceleration due to gravity} \] On the Moon, the acceleration due to gravity is one-sixth of that on Earth. Assume Earth's gravity \( g \) is approximately \( 9.8 \text{ m/s}^2 \). Therefore, Moon's gravity \( g_{\text{moon}} = \frac{9.8}{6} \text{ m/s}^2 = 1.63 \text{ m/s}^2 \).
04

Calculating the Object's Weight on the Moon

We use the formula for weight: \[ \text{Weight on Moon} = 6.0 \text{ kg} \times 1.63 \text{ m/s}^2 \] Perform the multiplication to find the weight: \[ \text{Weight on Moon} = 9.78 \text{ N} \] Thus, the weight of the object on the Moon is approximately 9.78 Newtons.

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Key Concepts

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

Understanding Mass vs Weight
It's important to understand that mass and weight are not the same thing. Mass refers to the amount of matter in an object. It's constant, meaning it doesn't change regardless of where you are in the universe. Whether you're on Earth, the Moon, or even in outer space, your mass remains unchanged. This is why, in the exercise's solution, the mass of the 6.0 kg object stays 6.0 kg when on the Moon.
Weight, on the other hand, is the force exerted by gravity on that mass. Because weight is dependent on gravitational pull, it can change depending on where you are. On Earth, you experience a certain weight due to Earth's gravity, while on the Moon, you experience less weight because the gravitational force is weaker there.
Key points to remember:
  • Mass is measured in kilograms (kg) and remains constant.
  • Weight is measured in Newtons (N) and is variable, depending on gravity.
Exploring Gravity on the Moon
Gravity on the Moon is significantly weaker than on Earth. While Earth's gravity is approximately 9.8 m/s虏, the Moon's gravity is only about one-sixth of that value, which is roughly 1.63 m/s虏. This difference is because the Moon is much smaller and less massive compared to Earth, resulting in weaker gravitational pull. Since gravity influences weight, this is why objects weigh less on the Moon, even though their mass remains unchanged. Understanding this difference in gravity helps us calculate weight accurately when transitioning between planets.
Some interesting facts about lunar gravity:
  • You could jump much higher on the Moon than on Earth due to its weaker gravity.
  • This is why lunar astronauts appear to move in slow motion in video footage.
Calculating Weight on Different Planets
To calculate the weight of an object on different planets, we use the universal weight equation: \[ \text{Weight} = \text{mass} \times \text{acceleration due to gravity} \]When you're dealing with objects on different celestial bodies, such as planets or moons, it's crucial to substitute the right gravity value for that location. In our example, to find the weight of the 6.0 kg object on the Moon, we use the Moon's gravitational acceleration of 1.63 m/s虏:
\[ \text{Weight on Moon} = 6.0 \text{ kg} \times 1.63 \text{ m/s}^2 = 9.78 \text{ N} \]
Here's a quick guide to calculating weights:
  • Identify the object's mass in kilograms.
  • Use the appropriate gravitational acceleration for the planet or moon. For example, Earth is 9.8 m/s虏, and Moon is 1.63 m/s虏.
  • Multiply these values to find the weight in Newtons.
This approach can be applied universally to calculate weights on any planet by using their respective gravitational values.

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Most popular questions from this chapter

A block that has a mass of \(2.0 \mathrm{~kg}\) and is \(10 \mathrm{~cm}\) wide on each side just begins to slide down an inclined plane with a \(30^{\circ}\) angle of incline ( \(\mathbf{r}\) Fig. 4.45 ). Another block of the same height and same material has base dimensions of \(20 \mathrm{~cm} \times 10 \mathrm{~cm}\) and thus a mass of \(4.0 \mathrm{~kg} .\) (a) At what critical angle will the more massive block start to slide down the plane? Why? (b) Estimate the coefficient of static friction between the block and the plane.

At the end of most landing runways in airports, an extension of the runway is constructed using a special substance called formcrete. Formcrete can support the weight of cars, but crumbles under the weight of airplanes to slow them down if they run off the end of a runway. If a plane of mass \(2.00 \times 10^{5} \mathrm{~kg}\) is to stop from a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) on a \(100-\mathrm{m}\) -long stretch of formcrete, what is the average force exerted on the plane by the formcrete?

A \(2.50-\mathrm{kg}\) block is placed on a rough surface inclined at \(30^{\circ} .\) The block is propelled and launched at a speed of \(1.60 \mathrm{~m} / \mathrm{s}\) down the incline and comes to rest after sliding \(1.10 \mathrm{~m} .\) (a) Draw the free-body diagram of the block while it is sliding. Also indicate your coordinate system axes. (b) Starting with Newton's second law applied along both axes of your coordinate system, use your free- body diagram to generate two equations. (c) Solve these equations for the coefficient of kinetic friction between the block and the incline surface. [Hint: You will need to first determine the block's acceleration.]

. \(\bullet\) A 1.5 -kg object moves up the \(y\) -axis at a constant speed. When it reaches the origin, the forces \(F_{1}=5.0 \mathrm{~N}\) at \(37^{\circ}\) above the \(+x\) -axis, \(F_{2}=2.5 \mathrm{~N}\) in the \(+x\) -direction, \(F_{3}=3.5 \mathrm{~N}\) at \(45^{\circ}\) below the \(-x\) -axis, and \(F_{4}=1.5 \mathrm{~N}\) in the \(-y\) -direction are applied to it. (a) Will the object continue to move along the \(y\) -axis? (b) If not, what simultaneously applied force will keep it moving along the \(y\) -axis at a constant speed?

During a daring rescue, a helicopter rescue squad initially accelerates a little girl (mass \(25.0 \mathrm{~kg}\) ) vertically off the roof of a burning building. They do this by dropping a rope down to her, which she holds on to as they pull her up. Neglect the mass of the rope. (a) What force causes the girl to accelerate vertically upward: (1) her weight; (2) the pull of the helicopter on the rope; (3) the pull of the girl on the rope; or (4) the pull of the rope on the girl? (b) Determine the pull of the rope (the tension) if she initially accelerates upward at \(0.750 \mathrm{~m} / \mathrm{s}^{2}\)
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