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Chapter 4: Problem 59

An object with mass \(m\) moves along the \(x\) -axis. Its position as a function of time is given by \(x(t)=A t-B t^{3},\) where \(A\) and \(B\) are constants. Calculate the net force on the object as a function of time.

Short Answer

Expert verified
The net force on the object is \(F(t) = -6mBt\).

Step by step solution

01

Understand Position Function

The position of the object is given by the function \(x(t) = A t - B t^3\). Here, \(A\) and \(B\) are constants, which indicates that the object's position changes over time based on this equation.
02

Calculate First Derivative for Velocity

Velocity \(v(t)\) is the first derivative of the position function with respect to time \(t\). We differentiate \(x(t) = A t - B t^3\) by applying the derivative: \[ v(t) = \frac{d}{dt}(A t - B t^3) = A - 3B t^2 \]Thus, the velocity function is \(v(t) = A - 3B t^2\).
03

Calculate Second Derivative for Acceleration

Acceleration \(a(t)\) is the derivative of the velocity function. Differentiate \(v(t) = A - 3B t^2\) with respect to \(t\):\[ a(t) = \frac{d}{dt}(A - 3B t^2) = -6B t \]The acceleration function is thus \(a(t) = -6B t\).
04

Use Newton's Second Law to Find Force

According to Newton's second law, force \(F(t)\) is the product of mass \(m\) and acceleration \(a(t)\):\[ F(t) = m a(t) = m (-6B t) \]Thus, the force as a function of time is \(F(t) = -6mBt\).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in classical mechanics that describes the relationship between motion and force. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. Mathematically, this is expressed as:\[ F = m imes a \]Where:
  • \(F\) stands for the net force acting on the object,
  • \(m\) is the object's mass,
  • \(a\) is the acceleration.
Applied to our problem, this law helps us find the net force on an object as its acceleration is known through differentiation (calculus). When you know acceleration from the position, you simply multiply by mass to find force. Newton's Second Law is the bridge that connects how an object's motion changes under the influence of known forces.
Velocity
Velocity is a key concept in understanding motion, describing both the speed and direction in which an object moves. In our exercise, the velocity function is derived as the first derivative of the position function with respect to time.From the position function \( x(t) = A t - B t^3 \), the velocity \( v(t) \) is:\[ v(t) = \frac{d}{dt}(A t - B t^3) = A - 3B t^2 \]This expression shows how the velocity of the object changes at each point in time.
  • At \(t = 0\), the initial velocity is \(A\).
  • Over time, the presence of the \(-3Bt^2\) term indicates that the velocity decreases, showing deceleration or changing direction based on \(B\)'s sign.
Understanding velocity is crucial since it's directly involved in determining acceleration, which then leads us to compute force using Newton's Second Law.
Acceleration
Acceleration describes how quickly velocity changes over time. It is calculated as the derivative of velocity. This means we look at how the rate of motion (velocity) itself changes.For velocity \( v(t) = A - 3B t^2 \), acceleration \( a(t) \) becomes:\[ a(t) = \frac{d}{dt}(A - 3B t^2) = -6B t \]This indicates that the acceleration depends linearly on time.
  • At \(t = 0\), the object has zero acceleration, implying any initial motion is due to initial velocity and not influenced by time yet.
  • As time progresses, acceleration either increases or decreases based on the sign of \(B\), affecting how velocity changes.
Understanding acceleration allows us to connect directly to Newton's Second Law, helping us to compute the force acting on the object.
Position Function
Position function is fundamental in describing an object's location at any given time. Given the equation \( x(t) = A t - B t^3 \), the position of the object depends on constants \(A\) and \(B\), and time \(t\).The function has several implications:
  • Constant \(A\) indicates the initial slope of the position when time starts, influencing initial motion.
  • Constant \(B\) impacts the curvature of the path, determining how the position changes more dramatically as time passes.
Breaking down position into velocity and acceleration through derivatives helps us further understand the nature of motion. The position function itself allows us to predict where an object will be at a particular time, setting the stage for deeper analysis with velocity, acceleration, and ultimately, using Newton's Second Law to find the force exerted on the object.

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Most popular questions from this chapter

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 \(x=\left(9.0 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(8.0 \times 10^{4} \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) 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 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 take out a 100 -g ball, release it from rest from a height of \(10.0 \mathrm{m},\) and measure that it takes 2.2 \(\mathrm{s}\) to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the \(100-\mathrm{g}\) ball weigh on the surface of Planet \(\mathrm{X} ?\)

An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spacesuit have a total mass of 105 \(\mathrm{kg}\) , while the mass of the cable is negligible. The mass of the spacecraft is \(9.05 \times 10^{4} \mathrm{kg} .\) The spacecraft is far from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) (a) What force does the cable exert on the astronaut? (b) since \(\Sigma \vec{\boldsymbol{F}}=m \vec{\boldsymbol{a}}\) , how can a "massless" \((m=0)\) cable exert a force? (c) What is the astronaut's acceleration? (d) What force does the cable exert on the spacecraft? (e) What is the acceleration of the spacecraft?

An object of mass \(m\) is at rest in equilibrium at the origin. At \(t=0\) a new force \(\vec{\boldsymbol{F}}(t)\) is applied that has components $$F_{x}(t)=k_{1}+k_{2} y \quad F_{y}(t)=k_{3} t$$ where \(k_{1}, k_{2},\) and \(k_{3}\) are constants. Calculate the position \(\vec{r}(t)\) and velocity \(\vec{v}(t)\) vectors as functions of time.

A crate with mass 32.5 \(\mathrm{kg}\) initially at rest on a warehouse floor is acted on by a net horizontal force of 140 \(\mathrm{N}\) . (a) What acceleration is produced? (b) How far does the crate travel in 10.0 \(\mathrm{s} ?\) (c) What is its speed at the end of 10.0 \(\mathrm{s} ?\)
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