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Chapter 3: Problem 53

An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon? (c) What will be its acceleration on the Moon under the action of a \(0.500-\mathrm{N}\) resultant force?

Short Answer

Expert verified
(a) 2.94 N; (b) 0.3 kg; (c) 1.67 m/s虏.

Step by step solution

01

Convert Mass to Kilograms

First, convert the mass of the object from grams to kilograms since the standard unit of mass in physics is kilograms. The object has a mass of 300 g. To convert grams to kilograms, divide by 1000:\[\text{mass in kg} = \frac{300\,g}{1000} = 0.3\,kg\]
02

Calculate Weight on Earth

Weight is the force due to gravity and can be calculated using the formula:\[\text{Weight} = \text{mass} \times \text{gravitational acceleration}\]On Earth, the gravitational acceleration \(g\) is approximately \(9.8\,m/s^2\).\[\text{Weight on Earth} = 0.3\,kg \times 9.8\,m/s^2 = 2.94\,N\]
03

Mass on the Moon

Mass remains constant regardless of location, whether on Earth or the Moon. Therefore, the mass of the object on the Moon is the same as its mass on Earth: Mass on the Moon = 0.3 kg.
04

Determine Acceleration on the Moon

Use Newton鈥檚 second law of motion, \(F = ma\), to calculate the acceleration, where \(F\) is the force applied and \(m\) is the mass of the object.Given that the force \(F\) is 0.500 N and the mass \(m\) on the Moon is 0.3 kg:\[a = \frac{F}{m} = \frac{0.500\,N}{0.3\,kg} = 1.67\,m/s^2\]

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Key Concepts

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

Mass Conversion
In physics, mass is typically measured in kilograms (kg), the standard unit for mass in the International System of Units (SI). When working with physics calculations, you might often need to convert mass from grams (g) to kilograms. This is a simple yet crucial step, as it ensures all measurements are consistent and compatible with other standard units.
To convert grams to kilograms, you simply divide the number of grams by 1000. This conversion factor is derived from the definition of the kilogram, whereby 1000 grams make up a kilogram. For example, an object with a mass of 300 grams has a mass of 0.3 kilograms when converted.
  • Grams to kilograms: Divide by 1000.
  • Consistency in units is critical for accurate calculations.
  • Effective unit conversion helps in applying physics principles globally.
Understanding how to convert mass accurately is essential for engaging with many physics problems, including those involving force and acceleration calculations.
Gravitational Force
Gravitational force is the attractive force between two masses. On Earth, this is sensed as weight, the force with which an object is pulled towards the center of the Earth. This force can be calculated using the formula: \[ \text{Weight} = \text{mass} \times \text{gravitational acceleration} \]Gravitational acceleration on Earth is approximately \(9.8 \,m/s^2\), a value that simplifies physics problems involving gravity.
To calculate an object's weight on Earth, multiply its mass in kilograms by the gravitational acceleration. For instance, an object with a mass of 0.3 kg has a weight of 2.94 N, showing how closely mass and gravitational force are related.
  • Weight = Mass x Gravitational Acceleration.
  • Earth's gravitational acceleration: \(9.8 \,m/s^2\).
  • Weight is a force measured in Newtons (N).
This concept of gravitational force is not limited to Earth. On the Moon, the gravitational acceleration is different, showing the variability of weight depending on location.
Newton's Second Law
Newton's second law provides a relationship between force, mass, and acceleration. It is often expressed as the equation: \[ F = ma \]Where \(F\) represents the force applied, \(m\) is the mass, and \(a\) is the acceleration.
To determine the acceleration experienced by an object when a force is applied, rearrange the formula to solve for \(a\): \[ a = \frac{F}{m} \]For example, if a force of 0.500 N is applied to a 0.3 kg object on the Moon, the resulting acceleration is \(1.67 \,m/s^2\).
  • Force, Mass, and Acceleration are interdependent.
  • Unit of force is Newtons (N).
  • Newton's second law is fundamental for understanding motion.
This principle is foundational in physics, providing a base to explore more complex interactions and motions.

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Most popular questions from this chapter

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0\) -kg package by a vertical string. What is the tension in the string during the accelerating process?

Just as her parachute opens, a 60 -kg parachutist is falling at a speed of \(50 \mathrm{~m} / \mathrm{s}\). After \(0.80 \mathrm{~s}\) has passed, the chute is fully open and her speed has dropped to \(12.0 \mathrm{~m} / \mathrm{s} .\) Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform.

Suppose, as depicted in Fig. 3-14, that a 70 -kg box is pulled by a 400-N force at an angle of \(30^{\circ}\) to the horizontal. The coefficient of kinetic friction is \(0.50\). Find the acceleration of the box. Fig. \(3-14\) Because the box does not move up or down, we have \(\sum F_{y}=m a_{y}=\) 0\. From Fig. \(3-14\), this equation is $$ +\uparrow \sum_{y}=F_{N}+200 \mathrm{~N}-m g=0 $$ But \(m g=(70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=686.7 \mathrm{~N}\), and it follows that \(F_{N}=\) \(486.7 \mathrm{~N}\) Next find the friction force acting on the box: $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.50)(486.7 \mathrm{~N})=243.4 \mathrm{~N} $$ Now write \(\sum F_{x}=m a_{x}\) for the box. It is $$ (346-243.4) \mathrm{N}=(70 \mathrm{~kg})\left(a_{X}\right) $$ from which \(a_{x}=1.466 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.5 \mathrm{~m} / \mathrm{s}^{2}\)

A 700 -N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of (a) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). \((a)\) How far will it slide before stopping? (b) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, \(\Sigma F=m a \quad\) becomes \(\quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a)\) from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(_{i x}\) \(=0.80 \mathrm{~m} / \mathrm{s}, f_{x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2} .\) Then \(v_{\mathrm{K}}^{2}-u_{u}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$
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