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Chapter 5: Problem 45

A car of mass \(1214.5 \mathrm{~kg}\) is moving at a speed of \(62.5 \mathrm{mph}\) when it misses a curve in the road and hits a bridge piling. If the car comes to rest in \(0.236 \mathrm{~s}\), how much average power (in watts) is expended in this interval?

Short Answer

Expert verified
Answer: The average power expended by the car during this interval is approximately 2.198 × 10^6 W.

Step by step solution

01

Convert speed from mph to m/s

First, we need to convert the speed of the car from miles per hour (mph) to meters per second (m/s). To do this, we can use the following conversion factor: 1 mph = 0.44704 m/s So, the speed in m/s is: \(62.5~\text{mph} \times 0.44704~\frac{\text{m}}{\text{s}} = 27.940~\frac{\text{m}}{\text{s}}\)
02

Calculate the kinetic energy

Now, we can find the kinetic energy of the car. The formula to calculate kinetic energy is: \(KE = \frac{1}{2}mv^2\) Where \(m\) is the mass of the car and \(v\) is its speed. Plugging in the values, we get: \(KE = \frac{1}{2} \times 1214.5~\text{kg} \times (27.940~\frac{\text{m}}{\text{s}})^2 = 518518.5326~\text{J}\) (Joules)
03

Calculate the work done

As the car comes to rest, its final kinetic energy becomes zero. So, the work done on the car will be equal to the initial kinetic energy (since work-energy theorem states that the work done on an object is equal to the change in its kinetic energy): \(W = KE_\text{initial} - KE_\text{final} = 518518.5326~\text{J} - 0~\text{J} = 518518.5326~\text{J}\)
04

Calculate the average power

Finally, we can calculate the average power expended by the car as it comes to a stop by dividing the work done by the time taken: \(P_\text{average} = \frac{W}{t} = \frac{518518.5326~\text{J}}{0.236~\text{s}} = 2197925.134~\text{W}\) Therefore, the average power expended by the car during this interval is approximately \(2197925.134~\text{W}\) or \(2.198 \times 10^6 ~\text{W}\).

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Most popular questions from this chapter

How much work is done when a \(75-\mathrm{kg}\) person climbs a flight of stairs \(10 \mathrm{~m}\) high at constant speed? a) \(7.35 \cdot 10^{5}\) J c) 75 e) 7350 J b) 750 J d) 7500 J

How much work is done against gravity in lifting a \(6.00-\mathrm{kg}\) weight through a distance of \(20.0 \mathrm{~cm} ?\)

A bullet moving at a speed of \(153 \mathrm{~m} / \mathrm{s}\) passes through a plank of wood. After passing through the plank, its speed is \(130 \mathrm{~m} / \mathrm{s}\). Another bullet, of the same mass and size but moving at \(92 \mathrm{~m} / \mathrm{s}\), passes through an identical plank. What will this second bullet's speed be after passing through the plank? Assume that the resistance offered by the plank is independent of the speed of the bullet.

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by \(5.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy, Calculate the original speeds of the two cars.

A force of \(5.00 \mathrm{~N}\) acts through a distance of \(12.0 \mathrm{~m}\) in the direction of the force. Find the work done.
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