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A 65 kg bicyclist rides his 7.8 kg bicycle with a speed of 15 m/s. How much work must be done by the brakes to bring the bike and rider to a stop?

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Final answer:

To stop the bicyclist and the bicycle, the brakes must perform work equal to the negative of the system's initial kinetic energy, which totals -8190 joules. This calculation uses the Work-Energy theorem, assuming no additional forces like friction are at play, and takes into account the combined mass of both rider and bicycle.

Step-by-step explanation:

The question pertains to the amount of work done by the brakes to bring a bicyclist and his bicycle to a stop. Given the mass of the bicyclist (65 kg) and the bicycle (7.8 kg) and the speed at which they are traveling (15 m/s), we can calculate this work using the Work-Energy theorem, which states that the work done is equal to the change in kinetic energy.

First, we calculate the total mass of the system (bicyclist plus bicycle) which is:

Total mass, m = mass of bicyclist + mass of bicycle = 65 kg + 7.8 kg = 72.8 kg

The kinetic energy (KE) of the bicyclist and bicycle is defined as:

KE = 1/2 * m * v^2

Where m is the mass and v is the velocity. Plugging in the known values:

KE = 1/2 * 72.8 kg * (15 m/s)^2

KE = 8190 J (joules)

Since the bicyclist is coming to a stop, the final kinetic energy is 0, so the work done by the brakes must be equal to the negative of the initial kinetic energy:

Work done by brakes, W = -KE

W = -8190 J

The negative sign indicates that the work done is in the opposite direction of the motion - in this case, it's the work required to decelerate the bicyclist and bicycle to a complete stop.

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