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A record company executive is on his way to a TV interview and is carrying a promotional CD in his briefcase. The mass of the briefcase and its contents is \(5.00 \mathrm{kg}\) The executive realizes that he is going to be late. Starting from rest, he starts to run, reaching a speed of $2.50 \mathrm{m} / \mathrm{s} .$ What is the work done by the executive on the briefcase during this time? Ignore air resistance.

Short Answer

Expert verified
Answer: The work done by the executive on the briefcase is 15.625 J.

Step by step solution

01

Find the initial kinetic energy of the briefcase.

Since the briefcase is initially at rest, its initial kinetic energy is zero. This can be derived from the formula for kinetic energy: $$KE = \frac{1}{2}m v^2$$ Where \(KE\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is the velocity of the object. Since the velocity is zero, the initial kinetic energy is also zero.
02

Find the final kinetic energy of the briefcase.

Now we will find the final kinetic energy of the briefcase when it reaches the speed of \(2.50 m/s\). We will use the kinetic energy formula, with the given mass of \(5.00 kg\) and the final velocity of \(2.50 m/s\): $$KE = \frac{1}{2}(5.00 kg)(2.50 m/s)^2$$ Calculate the final kinetic energy: $$KE = \frac{1}{2}(5.00 kg)(6.25 m^2/s^2)$$ $$KE = 15.625 J$$
03

Find the change in kinetic energy, which is equal to the work done.

Now we will find the change in kinetic energy, which is equal to the work done by the executive on the briefcase. To find the change in kinetic energy, we will subtract the initial kinetic energy (which is zero) from the final kinetic energy: $$W = KE_{final} - KE_{initial}$$ $$W = 15.625 J - 0 J$$ $$W = 15.625 J$$ So, the work done by the executive on the briefcase is \(15.625 J\).

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