91Ó°ÊÓ

Kinetic energy is the energy that an object possesses due to its motion. This concept is central when discussing mechanical systems and how forces affect the movement of objects.
When a particle or object moves, it gains or loses kinetic energy. The kinetic energy (KE) can be calculated using the formula:

• KE = \( \frac{1}{2} mv^2 \)

where \(m\) is the mass of the object, and \(v\) is its velocity.
In the exercise given, we have a particle with certain kinetic energies at two points: 20.0 J at \(x = 0\) meters and 11.0 J at \(x = 3.00\) meters. A change in kinetic energy implies that work has been done on the particle. This change can be calculated by the difference between the initial and final kinetic energies:

• \( \Delta KE = KE_{final} - KE_{initial} \)
• \( \Delta KE = 11.0 \text{ J} - 20.0 \text{ J} = -9.0 \text{ J} \)

This negative value indicates that the particle has lost energy while moving from the starting point to the endpoint.

The concept of work done by a force is crucial in understanding how energy is transferred to or from an object. Work is done when a force causes an object to move along a direction.
This exercise involves calculating the work done by the force \(\vec{F}\) as the particle moves from one position to another on the \(x\)-axis. The expression for work done, \(W\), when a force moves an object along the path is given by the integral over the path:

• \( W = \int F(x) \, dx \)

In our particular problem, the force is expressed as \( \vec{F} = (c x - 3.00 x^2) \hat{\mathrm{i}} \), and hence the work done by this force from \(x = 0\) to \(x = 3.00\) meters is written out as:

• \( W = \int_{0}^{3} (c x - 3.00 x^2) \, dx \)

This integral calculation determines the total work done by the varying force over the interval specified, which is crucial in linking the work done to the change in kinetic energy.

Integral calculus helps in determining the area under a curve which is precisely what is needed to calculate work done by variable forces. It enables us to evaluate physical quantities where simple formulas are not applicable.
In this scenario, the force \(F(x) = c x - 3.00 x^2\) is not constant, so we use integral calculus to compute the work done. We perform integration separately for each component of the force:

• \( \int_{0}^{3} c x \, dx = \left[ \frac{c}{2} x^2 \right]_{0}^{3} = \frac{9c}{2} \)
• \( \int_{0}^{3} 3.00 x^2 \, dx = \left[ x^3 \right]_{0}^{3} = 81 \)

Through separate integrations and algebraic manipulations, we find:

• \( W = \frac{9c}{2} - 81 \)

This allows us to set up the equation to solve for the unknown constant \(c\), linking the mathematics of integration to physical concepts of work and energy.