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A \(2.0 \mathrm{~kg}\) particle moves along an \(x\) axis, being propelled by a variable force directed along that axis. Its position is given by \(x=\) \(3.0 \mathrm{~m}+(4.0 \mathrm{~m} / \mathrm{s}) t+c t^{2}-\left(2.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}\), with \(x\) in meters and \(t\) in seconds. The factor \(c\) is a constant. At \(t=3.0 \mathrm{~s}\), the force on the particle has a magnitude of \(36 \mathrm{~N}\) and is in the negative direction of the axis. What is \(c\) ?

Short Answer

Expert verified
The value of c is 9 m/s².

Step by step solution

01

Determine the acceleration

Start with the kinematic equation for the particle's position: \[ x(t) = 3.0 + (4.0) t + c t^2 - (2.0) t^3 \] To find the acceleration, we first take the first derivative to find the velocity:\[ v(t) = \frac{dx}{dt} = 4.0 + 2ct - 6t^2 \] Then take the second derivative to find the acceleration:\[ a(t) = \frac{d^2x}{dt^2} = 2c - 12t \]
02

Use Newton's Second Law

According to Newton's Second Law, the force is given by \[ F = ma \] where \( m = 2.0 \) kg and \( F = 36 \) N (negative direction means the force is \(-36\) N). Thus, \[ -36 = 2.0 \times (2c - 12 \times 3) \]
03

Solve for c

Substitute \( t = 3.0 \) s into the second derivative and solve for \( c \):\[ -36 = 2.0 \times (2c - 36) \] Simplify and solve the equation:\[ -36 = 4c - 72 \]\[ -36 + 72 = 4c \]\[ 36 = 4c \]\[ c = 9 \]

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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 fundamental to understanding how objects move in response to forces. At its core, the law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is expressed as:
  • \( F = ma \)
In this equation:
  • \( F \) represents the net force applied to the object.
  • \( m \) is the mass of the object.
  • \( a \) is the acceleration produced.

The direction of the force is crucial. In the exercise, a force of \(-36 \text{ N}\) indicates that the particle is being slowed down because the force is acting in the opposite direction to the motion. When solving problems, understanding how these components interact helps predict motion behaviors.
Newton's Second Law also helps calculate the net force required to achieve a certain acceleration. By rearranging the formula, you can find unknowns in various scenarios by substituting known values. It is a powerful tool in physics for predicting and explaining the motion of objects under the influence of forces.
Acceleration
Acceleration is the rate at which an object changes its velocity. When an object speeds up, slows down, or changes direction, it experiences acceleration. In kinematic problems, calculating acceleration often involves taking the derivative of velocity.In the provided exercise, the initial position function is given by:
  • \( x(t) = 3 + 4t + ct^2 - 2t^3 \)
To find acceleration, you follow these steps:
  • First, derive the velocity function: \( v(t) = \frac{dx}{dt} = 4 + 2ct - 6t^2 \)
  • Then, derive the acceleration function: \( a(t) = \frac{d^2x}{dt^2} = 2c - 12t \)
Acceleration here is not constant but varies with time, largely due to the \(-12t\) term. This implies that as time \( t \) increases, the acceleration's contribution from this term becomes more significant, showing how the force affects the motion dynamically.
Particle Motion
Particle motion refers to the movement of a particle along a defined path due to forces acting upon it. In the context of kinematic equations, it often involves position, velocity, and acceleration.Consider the motion described by the given position function \( x(t) \). This function indicates how the particle's position changes over time, governed by both constant and variable terms:
  • \( 4t \) shows linear motion with a constant velocity.
  • \( ct^2 \) suggests an additional acceleration factor, where \( c \) is crucial in determining the exact curve of the path.
  • \(-2t^3\) represents a higher degree of motion complexity, affecting the particle's trajectory.
For any given problem, understanding these components helps one predict how the particle will move over time. Solving for these variables involves utilizing derivatives to find velocity and acceleration, then applying concepts such as Newton’s Second Law to interrelate force influences.

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Most popular questions from this chapter

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m}\). Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

(a) Figure \(5-47\) shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time \(t=0\), container 1 has mass \(1.30 \mathrm{~kg}\) and container 2 has mass \(2.80 \mathrm{~kg}\), but container 1 is losing mass (through a leak) at the constant rate of \(0.200 \mathrm{~kg} / \mathrm{s}\) At what rate is the acceleration magnitude of the containers changing at (a) \(t=0\) and (b) \(t=3.00 \mathrm{~s}\) ? (c) When does the acceleration reach its maximum value?

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 \(t=3.40 \mathrm{~s}\) ?

A \(52 \mathrm{~kg}\) circus performer is to slide down a rope that will break if the tension exceeds \(425 \mathrm{~N}\). (a) What happens if the performer hangs stationary on the rope? (b) At what magnitude of acceleration does the performer just avoid breaking the rope?

A hot-air balloon of mass \(M\) is descending vertically with downward acceleration of magnitude \(a\). How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude \(a\) ? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

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