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Chapter 5: Problem 25

Sunjamming. A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday cir- cumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and receives a push of \(20 \mathrm{~N}\). (a) What is the mag- nitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

Short Answer

Expert verified
(a) 0.0222 m/s虏 (b) 82,886.4 km (c) 1918.08 m/s

Step by step solution

01

Calculate the Acceleration

The acceleration of an object can be calculated using Newton's second law of motion: \( F = ma \), where \( F \) is the force applied, \( m \) is mass, and \( a \) is the acceleration. We can rearrange this formula to find \( a \) (acceleration) as \( a = \frac{F}{m} \). Given that \( F = 20 \) N and \( m = 900 \) kg, we can calculate the acceleration as follows:\[ a = \frac{20\, \text{N}}{900\, \text{kg}} = \frac{2}{90}\, \text{m/s}^2 = 0.0222\, \text{m/s}^2. \]
02

Calculate Distance Traveled in 1 Day

The distance traveled by an object starting from rest with constant acceleration can be calculated using the formula \( d = \frac{1}{2} a t^2 \), where \( d \) is distance, \( a \) is acceleration, and \( t \) is time in seconds. Since 1 day equals 86400 seconds, we substitute \( a = 0.0222\, \text{m/s}^2 \) and \( t = 86400\, \text{s} \) into the formula:\[ d = \frac{1}{2} \cdot 0.0222\, \text{m/s}^2 \cdot (86400\, \text{s})^2 = \frac{1}{2} \cdot 0.0222 \cdot 7464960000 \]\[ d \approx 82886400\, \text{meters} = 82886.4\, \text{km}. \]
03

Calculate Final Velocity After 1 Day

The final velocity of the spacecraft can be found using the formula for constant acceleration: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity (which is 0 since it starts from rest), \( a \) is acceleration, and \( t \) is time. Substituting \( a = 0.0222\, \text{m/s}^2 \) and \( t = 86400\, \text{s} \) gives:\[ v = 0 + 0.0222\, \text{m/s}^2 \times 86400\, \text{s} = 1918.08\, \text{m/s}. \]

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Key Concepts

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

Understanding Acceleration and Newton's Second Law
Acceleration is a core concept in physics that tells us how quickly the velocity of an object changes over time. Think of it as the rate of change of speed. To find acceleration, we make use of Newton's Second Law of Motion. This law is one of the most fundamental concepts in physics.
Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. We can write this relationship as:
  • \( F = ma \)
In this formula, \( F \) represents the force applied to the object (in newtons), \( m \) is the mass of the object (in kilograms), and \( a \) is the acceleration (in \( \text{m/s}^2 \)).
For example, if we know the mass of a sun yacht is 900 kg and it receives a push of 20 N, we rearrange the formula to solve for acceleration:
  • \( a = \frac{F}{m} \)
  • \( a = \frac{20 \text{ N}}{900 \text{ kg}} \approx 0.0222 \text{ m/s}^2 \)
This tells us how fast the sun yacht's speed is increasing every second.
Calculating the Distance Traveled During Acceleration
When an object starts moving from rest, and it has constant acceleration, the distance it travels over a specific period can be determined using a special formula. The formula is:
  • \( d = \frac{1}{2} a t^2 \)
Here, \( d \) is the distance traveled, \( a \) is the constant acceleration, and \( t \) is the time for which the object has been moving.
Suppose a spacecraft begins its journey and travels for one day. Since there are 86,400 seconds in a day, we substitute \( a = 0.0222 \text{ m/s}^2 \) and \( t = 86400 \text{ s} \):
  • \( d = \frac{1}{2} \, \times \, 0.0222 \, \text{ m/s}^2 \, \times \, (86400 \, \text{ s})^2 \)
  • \( d \approx 82886.4 \text{ km} \)
This means the craft will travel approximately 82,886.4 kilometers in one day due to the constant push of the sun's radiation.
Finding the Final Velocity After Constant Acceleration
To find out how fast an object is moving after a certain period of constant acceleration, we use a simple formula. This formula helps us calculate the final velocity:
  • \( v = u + at \)
In this equation, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time of travel.
If our spacecraft starts from rest, the initial velocity \( u \) is 0. After one day, with a constant acceleration of \( 0.0222 \text{ m/s}^2 \) and a time duration of 86,400 seconds, we get:
  • \( v = 0 + (0.0222 \text{ m/s}^2 \times 86400 \text{ s}) \)
  • \( v = 1918.08 \text{ m/s} \)
This means that after a day, the spacecraft will be moving at a speed of approximately 1918.08 meters per second.

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Most popular questions from this chapter

The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(90 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant deceleration.

Figure 5-37 shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time \(t=0\), container 1 has mass \(1.30 \mathrm{~kg}\) and container 2 has mass \(2.80 \mathrm{~kg}\), but container 1 is losing mass (through a leak) at the constant rate of \(0.200 \mathrm{~kg} / \mathrm{s}\). At what rate is the acceleration magnitude of the containers changing at (a) \(t=0\) and (b) \(t=3.00 \mathrm{~s}\) ? (c) When does the acceleration reach its maximum value?

Holding on to a towrope moving parallel to a frictionless ski slope, a \(45 \mathrm{~kg}\) skier is pulled up the slope, which is at an angle of \(8.0^{\circ}\) with the horizontal. What is the magnitude \(F_{\text {rope }}\) of the force on the skier from the rope when (a) the magnitude \(v\) of the skier's velocity is constant at \(2.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{b}) v=2.0 \mathrm{~m} / \mathrm{s}\) as \(v\) increases at a rate of \(0.10 \mathrm{~m} / \mathrm{s}^{2}\) ?

A lamp hangs vertically from a cord in a de- scending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the tension in the cord is \(93 \mathrm{~N}\), what is the lamp's mass? (b) What is the cord's tension Problem \(43 .\) when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) ?

In an \(x y\) plane, a \(0.450 \mathrm{~kg}\) object moves in such a way that \(x(t)=\) \(-16.0+3.00 t-5.00 t^{3}\) and \(y(t)=26.0+8.00 t-10.0 t^{2}\), where \(x\) and \(y\) are measured in meters and \(t\) in seconds. At \(t=0.800 \mathrm{~s}\), find (a) the magnitude and (b) the angle, relative to the positive direc- tion of the \(x\) axis, of the net force on the object, and (c) the angle of the object's travel direction.
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