91Ó°ÊÓ

The work-energy theorem is a fundamental principle in physics. It states that the work done on an object by the net force is equal to the change in the object’s kinetic energy.
In mathematical terms, the work-energy theorem is expressed as:
\[ W = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \]
Where \( W \) is the work done, \( m \) is the mass, \( v \) is the final velocity, and \( v_0 \) is the initial velocity.
This theorem helps us understand the relationship between force, work, and kinetic energy. In our exercise, the block starts from rest, so its initial kinetic energy is zero. By calculating the work done on the block, we directly find its kinetic energy as it moves.
If the force acting on the block is described by a function, then integrating this function over a given distance gives us the total work done and, thereby, the kinetic energy.

Work is the measure of energy transfer when a force moves an object over a distance. When the force varies, as in our exercise, we need to calculate work using integration.
The work done by a force \( \textbf{F}(x) \) over a displacement from \( x = 0 \) to \( x = 2.0 \, m \) is given by the integral:
\[ W = \int_{0}^{2} (2.5 \, N - x^2 \, N/m^2) \, dx \]
Breaking this down:

• The integral is the sum of infinitesimally small amounts of work, \( \textbf{F}(x) \, dx \).
• In our specific problem, \( \textbf{F}(x) = 2.5 - x^2 \).
• We find the total work by evaluating the integral over the specified range.

This use of integration for calculating work is crucial when dealing with variable forces. It makes sure all the infinitesimal changes in force over distance are accounted for, leading to accurate results.
After solving the integral, we found that the work done \( W \) was \( 2.33 \, J \). This tells us the kinetic energy of the block at that point.

Integration is a powerful tool in calculus to find areas under curves, which in physics can represent quantities like displacement, work, and kinetic energy.
The integral we computed in this exercise was:
\[ W = \int_{0}^{2} (2.5 - x^2) \, dx \]
We solve this by finding the antiderivative of the integrand:

• \( \int 2.5 \, dx = 2.5x \)
• \( \int -x^2 \, dx = -\frac{x^3}{3} \)

Next, we evaluate this from \( x = 0 \) to \( x = 2 \):
\[ W = [2.5x - \frac{x^3}{3}]^{2}_{0} = (2.5 \cdot 2 - \frac{2^3}{3}) - (2.5 \cdot 0 - \frac{0^3}{3}) \]
\[ W = 5 - \frac{8}{3} = \frac{15}{3} - \frac{8}{3} = \frac{7}{3} = 2.33 \, J \]
The result of the integral gives us the work done, which translates to kinetic energy.
Understanding how to set up and solve integrals allows us to handle various physics problems where quantities change over space or time.