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Chapter 5: Problem 10

A \(0.150 \mathrm{~kg}\) particle moves along an \(x\) axis according to \(x(t)=-13.00+2.00 t+4.00 t^{2}-3.00 t^{3}\), with \(x\) in meters and \(t\) in seconds. In unit-vector notation, what is the net force acting on the particle at \(t=3.40 \mathrm{~s}\) ?

Short Answer

Expert verified
The net force acting on the particle at \( t = 3.40 \mathrm{~s} \) is \(-7.98 \mathrm{~N}\) in the \( i \) direction.

Step by step solution

01

Understand the formula for acceleration

The net force acting on a particle can be determined using Newton's second law, \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration. To find acceleration, we need the second derivative of the position function with respect to time \( t \), i.e., \( a(t) = \frac{d^2x}{dt^2} \).
02

Find the first derivative, velocity

The velocity is the first derivative of the position function \( x(t) \) with respect to time \( t \). Calculate \( v(t) = \frac{dx}{dt} = \frac{d}{dt}[-13.00 + 2.00t + 4.00t^2 - 3.00t^3] \). This gives, \( v(t) = 2.00 + 8.00t - 9.00t^2 \).
03

Find the second derivative, acceleration

Now, differentiate the velocity function \( v(t) = 2.00 + 8.00t - 9.00t^2 \) with respect to \( t \) to find the acceleration: \( a(t) = \frac{dv}{dt} = 8.00 - 18.00t \).
04

Calculate acceleration at \( t = 3.40 \mathrm{~s} \)

Substitute \( t = 3.40 \) into the acceleration function \( a(t) = 8.00 - 18.00t \): \[ a(3.40) = 8.00 - 18.00 \times 3.40 = 8.00 - 61.20 = -53.20 \mathrm{~m/s^2} \].
05

Apply Newton's second law

Use Newton's second law \( F = ma \) to calculate the net force. Substitute \( m = 0.150 \mathrm{~kg} \) and \( a = -53.20 \mathrm{~m/s^2} \) into the equation: \[ F = 0.150 \times (-53.20) = -7.98 \mathrm{~N} \].

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Key Concepts

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

Kinematics
Kinematics is an essential branch of physics that deals with the motion of objects without considering the causes of motion. It provides tools to describe how objects move through space and time. In kinematics, we primarily focus on quantities like position, velocity, and acceleration.
  • Position: The location of an object in space relative to a reference point. In the given problem, the position of the particle is described by the function\(x(t) = -13.00 + 2.00t + 4.00t^2 - 3.00t^3\).
  • Velocity: The rate of change of position with respect to time. It's the first derivative of the position function.
  • Acceleration: The rate of change of velocity with respect to time, which is the second derivative of the position function.

In the context of Newton's second law exercise, understanding kinematics is crucial as it allows us to track the changing position over time and derive key quantities like velocity and acceleration steps.
Differentiation
Differentiation is a fundamental concept in calculus focusing on how quantities change. It is the process of finding the derivative, which represents the rate of change. In kinematics, differentiation helps us move from position to velocity and eventually to acceleration.

Step by Step Differentiation

The first and second derivatives are key to solving problems in motion and are very straightforward once the steps are clear.
  • First Derivative - Velocity: For a given position function, like\(x(t) = -13.00 + 2.00t + 4.00t^2 - 3.00t^3\), the first derivative with respect to time\(t\) is the velocity\(v(t)\). Here,\(v(t) = 2.00 + 8.00t - 9.00t^2\).
  • Second Derivative - Acceleration: Differentiating the velocity function gives the acceleration function. For\(v(t) = 2.00 + 8.00t - 9.00t^2\), the acceleration\(a(t)\) is\(8.00 - 18.00t\).

Differentiation is crucial in deriving these relationships necessary for using Newton鈥檚 second law effectively.
Acceleration
Acceleration is a vector quantity that indicates how an object's velocity changes over time. It provides insight into how quickly an object is speeding up, slowing down, or changing direction. In the realm of Newton鈥檚 second law, acceleration is pivotal as it connects with the force and mass of the object.
  • To find the acceleration, one must know the velocity or position as a function of time.
  • The acceleration given in our problem,\(a(t) = 8.00 - 18.00t\), describes how quickly the particle's velocity changes at any given point in time.

Calculating Specific Acceleration

To solve the exercise, determine the acceleration at a specific\(t\) value, which was\(t = 3.40\) seconds, as:\[a(3.40) = 8.00 - 61.20 = -53.20\, \mathrm{m/s^2}\].
Acceleration is central to applying Newton's second law\((F=ma)\) because it helps us understand the type and magnitude of force acting upon the particle at that point in time.

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Most popular questions from this chapter

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s}\), and its penetration depth was \(15 \mathrm{~mm}\). If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

A certain force gives an object of mass \(m_{1}\) an acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) and an object of mass \(m_{2}\) an acceleration of \(3.30\) \(\mathrm{m} / \mathrm{s}^{2} .\) What acceleration would the force give to an object of mass (a) \(m_{2}-m_{1}\) and (b) \(m_{2}+m_{1}\) ?

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude \(F=450 \mathrm{~N}\) on the rope, which is inclined at an upward angle \(\theta=38^{\circ}\) to the horizontal, and the floor exerts a horizontal force of magnitude \(f=125 \mathrm{~N}\) that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is \(310 \mathrm{~kg}\) and \((\mathrm{b})\) its weight is \(310 \mathrm{~N}\).
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