Question

A shot-putter accelerates a \(7.30-\mathrm{kg}\) shot from rest to \(14.0 \mathrm{~m} / \mathrm{s}\). If this motion takes \(2.00 \mathrm{~s}\), what average power was supplied?

Short answer

Solution: Step 1: The initial kinetic energy is 0, since the shot is initially at rest. The final kinetic energy can be calculated using the formula \(KE = \frac{1}{2}mv^2\), with \(m = 7.30 \mathrm{kg}\) and \(v = 14.0\mathrm{~m}/\mathrm{s}\). Step 2: The change in kinetic energy is equal to the final kinetic energy, as we subtract the initial kinetic energy (0) from the final kinetic energy found in Step 1. Step 3: The work done by the shot-putter is equal to the change in kinetic energy of the shot found in Step 2. Step 4: The average power supplied by the shot-putter is found by dividing the work done (from Step 3) by the time taken, which is 2.00 seconds.

Step-by-step solution

Find the initial and final kinetic energies of the shot

Since the shot is initially at rest, its initial kinetic energy is zero. We can then find the final kinetic energy using the formula for kinetic energy: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the shot and \(v\) is its final velocity. In this case, \(m = 7.30 \mathrm{kg}\) and \(v = 14.0\mathrm{~m}/\mathrm{s}\).

Calculate the change in kinetic energy

To find the change in kinetic energy, we subtract the initial kinetic energy from the final kinetic energy. Since the initial kinetic energy is zero, the change in kinetic energy is equal to the final kinetic energy that we found in Step 1.

Calculate the work done by the shot-putter

According to the work-energy theorem, the work done by the shot-putter is equal to the change in kinetic energy of the shot. So, the work done is equal to the value calculated in Step 2.

Calculate the average power supplied by the shot-putter

The average power supplied by the shot-putter is equal to the work done divided by the time taken. In this case, the time taken is 2.00 seconds. So, we can calculate the average power by dividing the work done (found in Step 3) by 2.00 seconds.

Key concepts

Kinetic Energy Calculation

Understanding kinetic energy is essential when analyzing moving objects and the forces that act upon them. Kinetic energy (\textsc{KE}), expressed in joules (J), quantifies the energy an object possesses due to its motion. The formula to calculate this energy is given by

\[\begin{equation}KE = \frac{1}{2}mv^2\text{,}\br\br where \(m\) is the mass of the object in kilograms (kg) and \(v\) is the velocity in meters per second (m/s). When a shot-putter accelerates a shot from rest to a given speed, the shot's kinetic energy increases from zero to a positive value, reflecting the work the shot-putter has done to achieve this motion.

In the given exercise, the mass of the shot is \(7.30 \mathrm{kg}\) and its final velocity is \(14.0 \mathrm{m/s}\). With the kinetic energy formula, the final kinetic energy is calculated, providing the basis to determine the work done and ultimately, the average power output during the throw.

Work-Energy Theorem

The work-energy theorem is a fundamental principle in physics that connects the dots between work and energy. It states that the work done on an object is equal to the change in its kinetic energy, or in mathematical terms:

\[\begin{equation}W = \Delta KE\text{,}\br\br where \(W\) is work and \(\Delta KE\) is the change in kinetic energy. If an object starts at rest, like our shot-put subject, the only kinetic energy to consider is the final kinetic energy after all the work has been done.

The theorem helps us solve problems by giving us a way to quantify the work involved in accelerating an object. If the kinetic energy increases, work has been done. Applying this principle to the exercise problem, we equate the final kinetic energy to the work done by the shot-putter. This connection is vital because it allows us to measure the shot-putter's efforts solely based on the motion of the shot.

Power and Work Relationship

After understanding work and kinetic energy, the next step is grasping the concept of power, which represents the rate at which work is done. The quicker the work is done, the more powerful the action. Power in physics, denoted as \(P\), is the work done (\textsc{W}) divided by the time (\textsc{t}) it takes to do that work:

\[\begin{equation}P = \frac{W}{t}\text{.}\br\br With SI units of power being watts (W), where 1 W equals 1 J/s, we are able to measure power as the energy transfer rate.

To find the average power supplied during the shot-put, we take the total work -- which from the work-energy theorem, we know is equal to the change in kinetic energy -- and divide it by the time it took for the shot to be accelerated. This is precisely what happens in Step 4 of the solution. By calculating the average power, we not only solve the physics problem at hand but also gauge the efficacy and energy transfer capabilities of the shot-putter's performance.