A machine does work at a rate given by \(P=c t^{2},\) where \(c=18 \mathrm{W} / \mathrm{s}^{2}\) and \(t\) is time. Find the work done between \(t=10 \mathrm{s}\) and \(t=20 \mathrm{s}\)

When discussing the work done by machines or any force-producing entity, it's essential to understand the relationship between power and time. Power is defined as the rate at which work is performed or energy is transferred. In the context of the given exercise, a machine does work with power changing over time according to the relationship given as P = ct^2, where c is a constant.

As time progresses, the power at which the machine performs work increases, since power is proportional to the square of time (t^2). It tells us how quickly the machine is working - the higher the power at any given moment, the more work is being done per unit of time. When we want to find the total work done over a period, we have to take into account how the power varies with time and not just consider a single moment.

To calculate the total work done by the machine, we cannot merely multiply the power at a single moment by the time period because the power is changing with time. Instead, we use integration, which is a mathematical tool for adding up an infinitely large number of infinitesimally small quantities.

Why Integration?

Since the power function P = ct^2 indicates that the power changes with each passing second, integration enables us to effectively 'sum' the work done at every moment during the interval from t = 10 s to t = 20 s. This continuous addition is what gives us the total work done over the time period.

Integrating the power function over time gives us the work because work is essentially the accumulation of energy transferred over the period, and since power is the rate of energy transfer, integration helps to account for the varying rates over time.

The definite integral is a powerful tool in physics because it helps us solve problems involving quantities that change over time. In the case of work done by a machine, the definite integral of the power function tells us the total work done between two points in time.

Practical Application

In our exercise, we used a definite integral to calculate the work W done by the machine from t = 10 s to t = 20 s by evaluating W = \[ (1/3)t^3 \]_{10}^{20}. The limits of the integral, 10 and 20, represent the start and end times of the work period. By computing the definite integral, we effectively measure the cumulative effect of the changing power output over the entire interval, thus obtaining the total work done.

Understanding how to apply definite integrals in physics is crucial for translating theoretical equations that describe how a system evolves into practical results, such as the work done or energy used over specific time intervals.