Question

A variable force is given by \(F(x)=A x^{6},\) where \(A=13.75 \mathrm{~N} / \mathrm{m}^{6} .\) This force acts on an object of mass \(3.433 \mathrm{~kg}\) that moves on a frictionless surface. Starting from rest, the object moves from \(x=1.105 \mathrm{~m}\) to a new position, \(x\). The object gains \(5.662 \cdot 10^{3} \mathrm{~J}\) of kinetic energy. What is the new position \(x ?\)

Short answer

Answer: The new position of the object is approximately x = 1.241 m.

Step-by-step solution

Write down the given force expression and initial position

The given force expression is \(F(x) = Ax^6\), where \(A = 13.75 N/m^6\). The initial position of the object is given as \(x = 1.105m\), and its mass is \(3.433kg\).

Write down the work-energy theorem expression and given kinetic energy

The work-energy theorem states that the work done (\(W\)) on an object is equal to the change in its kinetic energy (\(\Delta KE\)). Since the object starts from rest, the initial kinetic energy is zero. Therefore, this can be written as: \(W = \Delta KE\), where \(\Delta KE = 5.662 * 10^3 J\).

Calculate the work done by the variable force

Work done by a variable force can be found by integrating the force expression along the path of the object. In this case, integrating \(F(x)\) from the initial position to the new position \(x\). \(W = \int_{x_i}^{x_f} F(x) \, dx = \int_{1.105}^{x} (13.75x^6) \, dx\)

Compute the integration

Calculate the integral: $W= 13.75 \int_{1.105}^{x} x^6 \, dx = 13.75 \left[\frac{1}{7}x^7 \right]_{1.105}^{x} = 13.75 \left(\frac{1}{7}x^7 - \frac{1}{7}(1.105)^7\right) $

Equate work done to given kinetic energy and solve for x

Now, equate the expression for work done to the given kinetic energy and solve for \(x\): \(13.75\left(\frac{1}{7}x^7 - \frac{1}{7}(1.105)^7\right) = 5.662 * 10^3\) Divide both sides by \(13.75\) and then multiply by \(7\): \(x^7 - (1.105)^7 = \frac{7(5.662 * 10^3)}{13.75}\) Now, add \((1.105)^7\) to both sides and take the seventh root of the result to find \(x\): \(x = \sqrt[7]{(1.105)^7 + \frac{7(5.662 * 10^3)}{13.75}}\) Calculating this value, we get: \(x \approx 1.241m\) The new position of the object is approximately \(1.241m\).

Key concepts

Work-Energy Theorem

Understanding the work-energy theorem is crucial for solving variable force physics problems. Simply put, this theorem is a principle of mechanics that relates the work done by all forces acting on a particle to the change in its kinetic energy. In the context of our exercise, where a force expressed by a variable equation acts upon an object, it’s essential to realize that the total work done as the object moves from one position to another results in a change of the object’s kinetic energy.

When the object starts from rest, its initial kinetic energy is zero. This simplifies our problem, as the work done is directly equal to the object's final kinetic energy. To apply the theorem, we calculate the work by integrating the force over the distance the object traveled. This gives us a direct relationship, which in mathematical terms is expressed as
\[ W = \Delta KE \] where \(W\) is the work done, and \(\Delta KE\) is the change in kinetic energy. In the given problem, the change in kinetic energy has a specific value which allows us to set up an equation to find the unknown position.

Kinetic Energy

Kinetic energy is a form of energy that an object possesses due to its motion. It is dependent on two main factors: the mass of the object and the speed at which it is moving. The standard formula to calculate the kinetic energy (KE) of an object is
\[ KE = \frac{1}{2}mv^2 \] where \(m\) is the mass of the object and \(v\) is its velocity.

In the scenario outlined in the exercise, the object initially has no kinetic energy since it starts from rest. As the force does work on the object, kinetic energy is imparted, and the object accelerates. The gain in kinetic energy, given by the problem, allows us to work backwards to find out how far the object must have moved (the new position \(x\)) under the influence of the force.

One key point for students to grasp is that kinetic energy is a scalar quantity – it has magnitude but no direction – which simplifies calculations involving work, as we are working with absolute energy values and not vector quantities.

Integration in Physics

Integration is a fundamental concept in calculus and is particularly relevant in physics problems involving variable forces. It is a mathematical tool used to sum up many small factors to determine a total effect; for instance, finding the area under a curve in a graph that represents a physical quantity.

In our exercise, integration helps us calculate the work done by the variable force \(F(x) = Ax^6\) as the object moves from one position to another. We integrate the force function over the displacement, starting from the initial position \(x_i\) to the final position \(x_f\). The general work done by a variable force is typically computed using the integral:
\[ W = \int_{x_i}^{x_f} F(x) \, dx \] For each tiny movement \(dx\), we calculate the small amount of work done, and integrate these over the entire path to get the total work.

It's worth noting, for beginner students, that the integration symbol \(\int\) represents a summation of infinite small quantities, and the limits of the integration (in this case, from \(x_i\) to \(x_f\)) indicate the range over which we are summing these quantities. Understanding integration is key to tackling many physics problems, and in this context, it allows us to connect the variable force with the change in kinetic energy to determine the object’s final position.