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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 $$x=3.0 \mathrm{~m}+(4.0 \mathrm{~m} / \mathrm{s}) t+c t^{2}-\left(2.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \text { , }$$ 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 \(c ?\) (Include units.)

Short Answer

Expert verified
c = 9 m/s^2

Step by step solution

01

Write down the equation of motion

The equation of motion is given as \[ x(t) = 3.0 \, m + (4.0 \, m/s) \, t + c \, t^2 - (2.0 \, m/s^3) \, t^3 \]where x is in meters and t is in seconds.
02

Find the velocity

Differentiate the position function, \( x(t) \), with respect to time, \( t \), to get the velocity function. \[ v(t) = \frac{dx}{dt} = 4.0 \, m/s + 2c \, t - 3(2.0 \, m/s^3) \, t^2 \]Simplifying, \[ v(t) = 4.0 \, m/s + 2c \, t - 6.0 \, m/s^3 \, t^2 \]
03

Find the acceleration

Differentiate the velocity function, \( v(t) \), with respect to time, \( t \), to get the acceleration function. \[ a(t) = \frac{dv}{dt} = 2c - 2(6.0 \, m/s^3) \, t \]Simplifying, \[ a(t) = 2c - 12.0 \, m/s^3 \, t \]
04

Calculate the acceleration at t = 3.0 s

Substitute \( t = 3.0 \, s \) into the acceleration function. \[ a(3.0 \, s) = 2c - 12.0 \, m/s^3 \, (3.0 \, s) \]Simplifying, \[ a(3.0 \, s) = 2c - 36.0 \, m/s^2 \]
05

Use the given force to find c

The force on the particle at \( t = 3.0 \, s \) is given as \( 36 \, N \) in the negative direction along the x-axis. Use Newton's second law, \( F = ma \), where \( m = 2.0 \, kg \) and \( F = -36 \, N \). Substituting into the equation gives:\[ -36 \, N = 2.0 \, kg \cdot (2c - 36.0 \, m/s^2) \]Solve for \( c \):\[ -36 = 2(2c - 36) \]\[ -36 = 4c - 72 \]\[ 36 = 4c \]\[ c = 9 \, m/s^2 \]

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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 in understanding motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This can be represented as \(\text{F = ma}\). Here, \(\text{F}\) refers to force, \(\text{m}\) to mass, and \(\text{a}\) to acceleration. This law helps us understand how the velocity of an object changes when it is subjected to an external force. For instance, when a car accelerates, the engine produces a force. This force changes the car's speed according to its mass. The bigger the car's mass, the more force you need to change its velocity. This principle is also used in the problem where the force is given and the mass of the particle is known. By rearranging Newton’s second law formula, we can solve for unknowns such as acceleration or the constant 'c' in this scenario.
Differentiation in Physics
Differentiation is a powerful tool in physics used to determine rates of change. When we differentiate a position function with respect to time, we get the velocity function. Velocity tells us how fast an object changes its position. For example, for the given motion equation \(\text{x(t) = 3.0 m + (4.0 m/s) t + c t^2 - (2.0 m/s^3) t^3}\), differentiating with respect to time gives us the velocity function: \(\text{v(t) = 4.0 m/s + 2c t - 6.0 (m/s^3) t^2}\). Similarly, differentiating the velocity function gives us the acceleration function. Acceleration is the rate at which velocity changes over time. Therefore, differentiating not only helps us find velocity from position but also acceleration from velocity.
Acceleration Calculation
Acceleration can be calculated by differentiating the velocity function with respect to time. In this exercise, we initially determine the equations of motion and velocity. Then we differentiate the velocity function to obtain the acceleration function: \(\text{a(t) = 2c - 12.0 m/s^3 t}\). Next, we substitute the given time into this function. For example, at \(\text{t = 3.0 s}\), the acceleration is \(\text{a(3.0 s) = 2c - 36.0 m/s^2}\). To determine the value of 'c', we use Newton's second law, which relates force and acceleration. Given the force and mass of the particle, we solve for 'c' and find its value to be \(\text{9 m/s^2}\). Understanding how to derive and use these equations is crucial for solving many physics problems related to motion.

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Most popular questions from this chapter

A lamp hangs vertically from a cord in a descending elevator that slows down at \(2.4 \mathrm{~m} / \mathrm{s}^{2} .\) (a) If the tension in the cord is \(89 \mathrm{~N}\), what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) ?

An experimental rocket sled can be accelerated at a constant rate from rest to \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\). What is the magnitude of the required net force if the sled has a mass of \(500 \mathrm{~kg}\) ?

A firefighter with a weight of \(712 \mathrm{~N}\) slides down a vertical pole with an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\), directed downward. What are the magnitudes and directions of the vertical forces (a) on the firefighter from the pole and (b) on the pole from the firefighter?

When a nucleus captures a stray neutron, it must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming that the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\).

A Frenchman, filling out a form, writes \(" 78 \mathrm{~kg}\) " in the space marked poids (weight). However weight is a force and \(\mathrm{kg}\) is a mass unit. What do the French (among others) have in mind when they use mass to report their weight? Why don't they report their weight in newtons? How many newtons does this Frenchman weigh? How many pounds?

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