Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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/s²; acceleration at t = 0 is 2.7 m/s², at t = 3 s is 16.86 m/s²; in space at t = 3 s is 26.56 m/s².

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:\[\text{Acceleration} = \frac{F}{m} = \frac{212.5}{8} = 26.56 \, \text{m/s}^2\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 14.0 N. (a) What acceleration is produced? (b) How far does the crate travel in 10.0 s? (c) What is its speed at the end of 10.0 s?

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F =\) 40.0 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.

A hockey puck with mass 0.160 kg is at rest at the origin (\(x =\) 0) on the horizontal, frictionless surface of the rink. At time \(t =\) 0 a player applies a force of 0.250 N to the puck, parallel to the \(x\)-axis; she continues to apply this force until \(t =\) 2.00 s. (a) What are the position and speed of the puck at \(t =\) 2.00 s? (b) If the same force is again applied at \(t =\) 5.00 s, what are the position and speed of the puck at \(t =\) 7.00 s?

To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by \(x =\) (9.0 \(\times\) 10\(^3\) m/s\(^2)t^2\) - (8.0 \(\times\) 10\(^4\) m/s\(^3)t^3\). The object leaves the end of the barrel at \(t =\) 0.025 s. (a) How long must the gun barrel be? (b) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a 1.50-kg object at (i) \(t =\) 0 and (ii) \(t =\) 0.025 s?

You have just landed on Planet X. You release a 100-g ball from rest from a height of 10.0 m and measure that it takes 3.40 s to reach the ground. Ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free