A small 8.00 8.00 -kg rocket burns fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation F = A + B t 2 F=A+Bt^2 . Measurements show that at t = 0 t = 0 , the force is 100.0 100.0 N, and at the end of the first 2.00 2.00 s, it is 150.0 150.0 N. Find the constants A A and B B , including their SI units.

Hey everyone today, we're dealing with a problem regarding forces. Now, at first glance, this looks to be a pretty lengthy problem. We have a formula. We have a bunch of different data sets, right? But um let's take a step back. Just sort of analyze this. So we're given a few key specific parts of info. We have a, excuse me, We have a spacecraft with a massive 12 kg. The forces acting upon the spacecraft obey relations. Such that F at specific time is equal to a plus B, times times squared. Now, at different instants of time, The acceleration is measured, or rather the force at that time is measured. So at time zero we have a force of 60 newtons and at time 2.6 seconds We have a force of 112 newtons. So with this we're being asked to determine the values of both A and B and S. I. Units. Now, let's go ahead and work through this. So let's start off very simply we have our we have our formula or relation which states that if a time T is equal to a plus B T squared, So at time is equal to zero when T is equal to zero seconds, F is equal to 60 Newtons. So substituting that in, we get 16 newtons is equal to a plus B into zero. So this will automatically become zero and we're left with a is equal to 16 newtons. So that is our first part. That is the value of A. But to find out the value of B, we need to use this newly acquired value of A. Again using the second set of instantaneous uh force and time. So let me write this in blue. We have F. Of T equal to A plus B into T squared. And we know that at time is equal to 2.60 seconds f is equal to 112 Newtons. And we know that A. Is 16 newtons. So substituting these values, you go 112 newtons is equal to Eaton's plus B into T squared. And now T squared is 2.6 seconds squared. So This means we have newtons is equal to 6.76 seconds squared into B. And therefore B is equal to 52 newtons divided by 6.76 seconds squared, which is equal to 7.69 newtons per second squared. So that is our answer A has a value of 60 newtons and B has a value of six points or 7.69 newtons per second squared. In other words, it is answer choice. Excuse me, it is an answer choice. D. I hope this helps. And I look forward to seeing you all in the next one

Table of contents

Skip topic navigation

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

6. Intro to Forces (Dynamics)

Newton's First & Second Laws

Physics

6. Intro to Forces (Dynamics)

Newton's First & Second Laws

Previous problemNext problem

3:53 minutes

Problem 15a

Textbook Question

A small $8.00$ -kg rocket burns fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation $F = A + B t^{2}$ . Measurements show that at $t equals 0$ , the force is $100.0$ N, and at the end of the first $2.00$ s, it is $150.0$ N. Find the constants $A$ and $B$ , including their SI units.

Verified step by step guidance

Step 1: Start by understanding the given force equation F = A + Bt², where A and B are constants to be determined. The force is time-dependent, and we are given two specific values of force at different times: F = 100.0 N at t = 0 and F = 150.0 N at t = 2.00 s.

Step 2: Substitute the first condition (t = 0, F = 100.0 N) into the equation F = A + Bt². Since t = 0, the term Bt² becomes zero, leaving F = A. Therefore, A = 100.0 N. This gives us the value of A in SI units (Newtons).

Step 3: Substitute the second condition (t = 2.00 s, F = 150.0 N) into the equation F = A + Bt². Using the value of A found in Step 2, the equation becomes 150.0 = 100.0 + B(2.00)². Simplify this to solve for B.

Step 4: Rearrange the equation from Step 3 to isolate B. Subtract 100.0 from both sides to get 50.0 = B(4.00). Then divide both sides by 4.00 to find B. The units of B can be determined by analyzing the equation: since F is in Newtons and t² is in seconds², B must have units of N/s².

Step 5: Verify the units of A and B. A is a constant force, so its units are Newtons (N). B is associated with the time-squared term, so its units are N/s². This ensures consistency with the SI unit system.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This law is often expressed with the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. Understanding this principle is crucial for analyzing the forces acting on the rocket and determining its motion.

Force Equation

The force exerted on the rocket is described by the equation F = A + Bt², where A and B are constants that define how the force changes over time. This equation indicates that the force is not constant but varies with the square of time, which is essential for calculating the values of A and B based on the given conditions at specific time intervals.

SI Units and Dimensional Analysis

SI units are the standard units of measurement used in science, including kilograms (kg) for mass, newtons (N) for force, and seconds (s) for time. Dimensional analysis involves checking the consistency of units in equations to ensure they are valid. In this problem, understanding SI units is necessary for correctly identifying the units of the constants A and B in the force equation.

Watch next

Master Intro to Forces & Newton's Second Law with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Multiple Choice

100 kg

rock is in space, far from any planets, stars or black holes. It is moving in a straight line at a constant

10,000 m / s

relative to our sun. What total force is required to keep it moving that fast?

1371

views

Multiple Choice

An elevator is going up at a constant speed in a very tall building. Assume the only forces acting on the elevator are a downward weight force and the force of a cable pulling the elevator up. Ignore air resistance and friction. How does the magnitude of the weight force compare to the magnitude of the force of the cable?

696

views

Multiple Choice

10 N

horizontal force is applied to a

10 kg

box causing it to speed up. The force of kinetic friction is

5 N

. Which choice best describes the subsequent motion of the box?

716

views

Textbook Question

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

A $4.50$ -kg experimental cart undergoes an acceleration in a straight line (the $x$ -axis). The graph in Fig. E $4.13$ shows this acceleration as a function of time. During what times is the net force on the cart a constant?

798

views

Textbook Question

A $4.50$ -kg experimental cart undergoes an acceleration in a straight line (the $x$ -axis). The graph in Fig. E $4.13$ shows this acceleration as a function of time. Find the maximum net force on this cart. When does this maximum force occur?

2242

views

Textbook Question

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. If the same force is again applied at $t equals 5.00$ s, what are the position and speed of the puck at $t equals 7.00$ s?

2185

views

Textbook Question

A dockworker applies a constant horizontal force of $80.0$ N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves $11.0$ m in $5.00$ s. What is the mass of the block of ice?

1294

views

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap