/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A small 8.00-kg rocket burns fue... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 15

A small 8.00-kg rocket burns fuel that exerts a timevarying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation \(F = A + Bt^2\). Measurements show that at \(t\) = 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N. (a) Find the constants \(A\) and \(B\), including their SI units. (b) Find the \(net\) force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

Short Answer

Expert verified
A = 100 N, B = 12.5 N/s虏; acceleration at t = 0 is 2.7 m/s虏, at t = 3 s is 16.86 m/s虏; in space at t = 3 s is 26.56 m/s虏.

Step by step solution

01

Identify Given Variables and Equations

We need to find constants \( A \) and \( B \) given the force equation \( F = A + Bt^2 \). The force measurements are \( F(0) = 100.0 \) N and \( F(2) = 150.0 \) N.
02

Solve for Constant A

Using the equation for \( t = 0 \), substitute values into the equation: \[F = A + B(0)^2 = 100 \, \text{N}\]This simplifies to \( A = 100.0 \) N.
03

Solve for Constant B

Using \( t = 2 \, \text{s} \), substitute into the equation:\[F = A + B(2)^2 = 150 \, \text{N}\]Substitute \( A = 100 \) N:\[150 = 100 + 4B\]Solve for \( B \):\[B = \frac{50}{4} = 12.5 \, \text{N/s}^2\]
04

Calculate Net Force and Acceleration at t=0

At \( t = 0 \), with gravity acting downwards:\[\text{Net force} = F - mg = 100 - (8 \times 9.8) = 21.6 \, \text{N}\]Acceleration \( a \) is given by:\[a = \frac{\text{Net Force}}{m} = \frac{21.6}{8} = 2.7 \, \text{m/s}^2\]
05

Calculate Net Force and Acceleration at t=3

At \( t = 3 \, \text{s} \):\[F = 100 + 12.5 \times (3)^2 = 100 + 112.5 = 212.5 \, \text{N}\]Net force:\[\text{Net force} = 212.5 - (8 \times 9.8) = 134.9 \, \text{N}\]Acceleration:\[a = \frac{134.9}{8} = 16.86 \, \text{m/s}^2\]
06

Calculate Acceleration in Space at t=3

In space, only the force \( F \) affects the rocket:\[\text{Acceleration} = \frac{F}{m} = \frac{212.5}{8} = 26.56 \, \text{m/s}^2\]

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Key Concepts

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

Time-Varying Force
In our study of Newton's Second Law, a key aspect is understanding how forces change over time. This is what we call a time-varying force. In our exercise, the force exerted by the rocket fuel changes as a function of time, expressed by the equation \( F = A + Bt^2 \). Here, both \( A \) and \( B \) are constants that determine how the force behaves at different moments.

**Understanding Time-dependence**:
  • When \( t = 0 \), the force is simply \( A \).
  • As time progresses, the term \( Bt^2 \) grows, indicating the force increases with time.
The force isn't static; instead, it adapts over time, which is crucial for scenarios like rocket launches, where varied force application affects acceleration and trajectory.
Net Force Calculation
To calculate the net force acting on the rocket, we must account for all existing forces. This typically involves both the force applied by the rocket's fuel and gravitational force. The equation to find the net force is:
  • Net force \( = F - mg \)
Here, \( F \) is the upward force from the rocket, \( m \) is the rocket's mass, and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.

**Net Force in Different Scenarios**:
  • At \( t = 0 \), the net force is just the initial force minus the gravitational force.
  • As time increases, \( F \) increases because of the term \( Bt^2 \).
  • Thus, the net force also increases, affecting acceleration.
By understanding the forces at play, we can predict the motion of the rocket at different time intervals.
Rocket Propulsion
Rocket propulsion is a fascinating application of Newton's Second Law, and it centers around the explosion of fuel to exert force. The time-varying nature of this force is crucial, as it influences how fast and how far a rocket can travel.

**Key Points in Rocket Propulsion**:
  • Rockets utilize the force generated from burning fuel to push against the mass of the rocket.
  • This propulsion is a direct application of Newton's law: the action of expelled gases results in a reaction force propelling the rocket.
  • The higher the net force, the greater the acceleration (\( a = \frac{ ext{net force}}{m} \)), meaning faster propulsion into space.
  • In space, without gravitational forces, the only force acting is the propulsion force, maximizing acceleration.
These principles not only explain how rockets escape Earth's gravity, but also how they maneuver in the vacuum of space.

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Most popular questions from this chapter

An advertisement claims that a particular automobile can "stop on a dime." What net force would be necessary to stop a 850-kg automobile traveling initially at 45.0 km/h in a distance equal to the diameter of a dime, 1.8 cm?

Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton's laws and the results of part (b) to calculate the average force he applied to the ground.

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (\(\textbf{Fig. P4.39}\)). A woman wearing golf shoes (for traction) pulls horizontally on the 6.00-kg crate with a force \(F\) that gives the crate an acceleration of 2.50 m/s\(^2\). (a) What is the acceleration of the 4.00-kg crate? (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton's second law to find the tension \(T\) in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, \(T\) or \(F\)? (d) Use part (c) and Newton's second law to calculate the magnitude of \(F\).

(a) An ordinary flea has a mass of 210 \(\mu\)g. How many newtons does it weigh? (b) The mass of a typical froghopper is 12.3 mg. How many newtons does it weigh? (c) A house cat typically weighs 45 N. How many pounds does it weigh, and what is its mass in kilograms?

Crates \(A\) and \(B\) sit at rest side by side on a frictionless horizontal surface. They have masses \(m_A\) and \(m_B\), respectively. When a horizontal force \(\vec{F}\) is applied to crate \(A\), the two crates move off to the right. (a) Draw clearly labeled free-body diagrams for crate \(A\) and for crate \(B\). Indicate which pairs of forces, if any, are third-law action-reaction pairs. (b) If the magnitude of \(\vec{F}\)is less than the total weight of the two crates, will it cause the crates to move? Explain.
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