Question

A particle moves parallel to the $x$ -axis. The net force on the particle increases with $x$ according to the formula $F_x$ $=(120\mathrm{N}/\mathrm{m})x,$ where the force is in newtons when $x$ is in meters. How much work does this force do on the particle as it moves from $x=0$ to $x=0.50\mathrm{m}?$ a) 7.5 J c) $30J$ e) 120 J b) 15 J d) 60

Short answer

Answer: The work done is 15 J.

Step-by-step solution

Recall the formula for work done by a variable force

The work done by a variable force $F_x$ when moving from position $x_1$ to $x_2$ is given by: $$W={\int }_{x_1}^{x_2}{F}_{x}dx$$ In this case, the force function is $F_x=120x$, and we are asked to find the work done as the particle moves from $x=0$ to $x=0.50m$.

Set up the integral

Using the given force function and limits, we can set up the integral as follows: $$W={\int }_0^{0.50}120xdx$$

Evaluate the integral

Integrating the force function with respect to $x$, we get: $$W=60x^2{|}_0^{0.50}$$

Compute the work

Now, apply the limits to find the work done: $$W=60({0.50}^2)-60(0^2)$$ $$W=60(0.25)$$ $$W=15J$$ So, the correct answer is 15 J.

Key concepts

Variable Force

When dealing with physical systems, a force that changes in magnitude or direction over time or position is called a variable force. Unlike constant forces, variable forces require different methods to analyze their effects, especially when calculating work done.
In our exercise, the force acting on the particle is defined as a function of position, given by the equation $F_x=120x$. Here:

• As the particle moves along the $x$-axis, the force changes proportionally with the position $x$.
• This means the further the particle is from the origin, the larger the force is applied to it.
• This changing force characteristic is what makes it variable as opposed to constant.

Understanding variable forces is crucial because they often appear in real-world problems. Changes in force occur in areas like springs, gravitational fields, and electrical forces, and grasping how they work is key to analyzing these systems.

Integral Calculus

Integral calculus helps us in finding the total accumulation of a quantity—like area, volumes, or, in our case, work, when dealing with variable forces. To determine the work done by a force on a particle, we use the concept of integration.
The formula for the work done by a variable force is:
$$W={\int }_{x_1}^{x_2}{F}_{x}dx$$
In the given problem, the integration calculates the work done by the force $F_x=120x$ as the particle moves from one point to another.

• The limits of integration $x_1$ and $x_2$ are the start and end positions: from $0$ to $0.50\text{m}$.
• The force function $120x$ is set inside the integral to compute the variable response across the interval.

Thus, integrating means summing up all the infinitesimally small work contributions across the movement path, yielding a total work done of 15 J.

Physics Problems

Physics problems like the one we solved require not only a clear understanding of mathematical concepts but also an ability to apply them to real-world situations. Here's how you can approach such problems:
First, understand the scenario. In our exercise, the particle is moving under the influence of a variable force along the $x$-axis.

• The mathematical relationship $F_x=120x$ describes how force varies with position.
• Recognize this as a typical physics setup where variable forces are at play.

Next, recall the appropriate physics laws and formulas. For work done by variable force, remember to use integral calculus.
Finally, set up the mathematical problem correctly by defining your variables and integrating properly. In this case, we needed to evaluate an integral to find the work from point $x=0$ to $x=0.50\text{m}$.
By practicing and understanding these steps, solving physics problems becomes an easier task and can help demystify complex concepts.