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A single force acts on a \(3.0 \mathrm{~kg}\) particle-like object whose position is given by \(x=3.0 t-4.0 t^{2}+1.0 t^{3}\), with \(x\) in meters and \(t\) in seconds. Find the work done by the force from \(t=0\) to \(t=4.0 \mathrm{~s}\).

Short Answer

Expert verified
The work done by the force is 528.0 Joules from \(t=0\) to \(t=4.0\) seconds.

Step by step solution

01

Find the velocity as a function of time

The velocity of an object is the derivative of its position with respect to time. Given the position equation \(x = 3.0t - 4.0t^2 + 1.0t^3\), calculate the velocity function:\[v(t) = \frac{d}{dt}(3.0t - 4.0t^2 + 1.0t^3) = 3.0 - 8.0t + 3.0t^2.\]
02

Find the acceleration as a function of time

Acceleration is the derivative of velocity with respect to time. Use the velocity function found in Step 1 to calculate the acceleration:\[a(t) = \frac{d}{dt}(3.0 - 8.0t + 3.0t^2) = -8.0 + 6.0t.\]
03

Calculate force as a function of time

Force is mass times acceleration. Given the mass \(m = 3.0 \text{ kg}\), use this to find the force function:\[F(t) = m \cdot a(t) = 3.0 \cdot (-8.0 + 6.0t) = -24.0 + 18.0t.\]
04

Determine the work done using the force

The work done by a force is the integral of force with respect to displacement. First, find the displacement function:\(x(t)\). Now, we need to calculate:\[W = \int_{t=0}^{t=4.0} F(t) \cdot v(t) \, dt.\]Substituting the expressions for \(F(t)\) and \(v(t)\): \[W = \int_{0}^{4.0} (-24.0 + 18.0t)(3.0 - 8.0t + 3.0t^2) \, dt.\]Evaluate this integral.
05

Evaluate the integral to find work done

To compute the integral, expand the expression:\[(-24.0 + 18.0t)(3.0 - 8.0t + 3.0t^2) = -72.0 + 192.0t - 72.0t^2 + 54.0t - 144.0t^2 + 54.0t^3 = -72.0 + 246.0t - 216.0t^2 + 54.0t^3.\]Then, compute the integral:\[W = \left[ -72.0t + \frac{246.0}{2}t^2 - \frac{216.0}{3}t^3 + \frac{54.0}{4}t^4 \right]_0^4 = \left[ -72.0t + 123.0t^2 - 72.0t^3 + 13.5t^4 \right]_0^4.\]Substitute \(t = 4.0\) and \(t = 0\) to find:\[W = \left(-72.0 \times 4.0 + 123.0 \times 16 - 72.0 \times 64.0 + 13.5 \times 256.0 \right)= -288.0 + 1968.0 - 4608.0 + 3456.0 = 528.0 \, \text{Joules}.\]
06

Conclusion: State the final answer

The work done by the force from \(t=0\) to \(t=4.0 \text{ s}\) is \(528.0 \text{ Joules}.\)

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Key Concepts

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

Kinematics
When studying motion, kinematics is a fundamental concept. It's all about describing how objects move, focusing on their position, velocity, and acceleration. In the provided exercise, we start with an expression for an object's position as a function of time. Specifically, its position is given by the equation \(x = 3.0t - 4.0t^2 + 1.0t^3\). This tells us where the object is located at any given time \(t\).

To further understand its motion, we derive the velocity of the object by taking the derivative of the position function. The result, \(v(t) = 3.0 - 8.0t + 3.0t^2\), describes how fast the object is moving at any given point over time. The change in velocity over time gives us acceleration, derived as \(a(t) = -8.0 + 6.0t\). This entire process is core to understanding how the parameters of motion are interconnected.
Force Calculation
Force plays a critical role in the dynamics of motion. According to Newton's second law, force is calculated as the product of mass and acceleration. In the given scenario, the mass \(m\) of the object is \(3.0 \text{ kg}\).

We already found the acceleration as a function of time, \(a(t) = -8.0 + 6.0t\). By multiplying the mass by the acceleration, we derive the force function: \(F(t) = m \cdot a(t) = 3.0 \cdot (-8.0 + 6.0t) = -24.0 + 18.0t\). This relation reveals how the force acting on an object varies with time as it moves.
Integral Calculus
Integral calculus is crucial for finding work done on an object when the force varies over a path. The work done by a force is the integral of the force component along the direction of displacement.

In our problem, after determining that \(F(t) = -24.0 + 18.0t\) and \(v(t) = 3.0 - 8.0t + 3.0t^2\), we integrate the product of force and velocity over the time interval from \(t=0\) to \(t=4.0\):
  • Expand the expression \((-24.0 + 18.0t)(3.0 - 8.0t + 3.0t^2)\).
  • Compute the integral of this expanded expression over the specified time interval.
Performing these steps gives us a deeper understanding of how to use integration to evaluate physical quantities in varying conditions.
Motion Analysis
Analyzing motion involves observing how objects accelerate, how forces change, and how they work together to produce motion. In the exercise, we analyze the motion by breaking down the components that affect a particle's movement.

We began by investigating the kinematic variables—position, velocity, and acceleration. This allows us to describe how an object's motion evolves over time.

Next, we calculated the force using the mass and acceleration, which shows the direct influence of force on motion. Finally, by employing integral calculus, we measured the work done by the varying force. This comprehensive analysis reveals the interplay between different physical quantities that govern motion, providing valuable insights into the mechanics of moving objects.

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Most popular questions from this chapter

A coin slides over a frictionless plane and across an \(x y\) coordinate system from the origin to a point with \(x y\) coordinates \((3.0 \mathrm{~m}, 4.0 \mathrm{~m})\) while a constant force acts on it. The force has magnitude \(2.0 \mathrm{~N}\) and is directed at a counterclockwise angle of \(100^{\circ}\) from the positive direction of the \(x\) axis. How much work is done by the force on the coin during the displacement?

A machine carries a \(4.0 \mathrm{~kg}\) package from an initial position of \(\vec{d}_{i}=(0.50 \mathrm{~m}) \hat{\mathrm{i}}+(0.75 \mathrm{~m}) \hat{\mathrm{j}}+(0.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=0\) to a final posi- tion of \(\vec{d}_{f}=(7.50 \mathrm{~m}) \hat{\mathrm{i}}+(12.0 \mathrm{~m}) \hat{\mathrm{j}}+(7.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=12 \mathrm{~s}\). The constant force applied by the machine on the package is \(\vec{F}=(2.00 \mathrm{~N}) \hat{\mathrm{i}}+(4.00 \mathrm{~N}) \hat{\mathrm{j}}+(6.00 \mathrm{~N}) \hat{\mathrm{k}}\). For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package.

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right) .\) Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).

A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

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