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Q45 PE

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College Physics (Urone)
Found in: Page 262
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

(a) What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00 s?

(b) Considering the amount of power generated, do you think a well-trained athlete could do this repetitively for long periods of time?

(a) The power output of the sprinter is \(1166.67{\rm{ W}}\) or \(1.56{\rm{ hp}}\).

(b) Probably not, since a human is much less powerful.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Work-energy theorem

According to work-energy theorem, when some work done on a moving body will result in change in the kinetic energy of the body. In other words, the change in kinetic energy of the body equals to the amount of the work done on it.

Power output in watts and horsepower

(a)

According to work-energy theorem, the work done by the sprinter is,

\(\begin{aligned}W &= \Delta {\rm{KE}}\\ &= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\end{aligned}\)

Here, m is the mass of the sprinter \(\left( {m = 70.0{\rm{ kg}}} \right)\), \({v_f}\) is the final velocity of the sprinter \(\left( {{v_f} = 10{\rm{ m}}/{\rm{s}}} \right)\), and \({v_i}\) is the initial velocity of the sprinter (\({v_i} = 0\) as the sprinter starts from rest).

Putting all known values,

\(\begin{aligned}W &= \frac{1}{2} \times \left( {70.0{\rm{ kg}}} \right) \times {\left( {10.0{\rm{ m}}/{\rm{s}}} \right)^2} - \frac{1}{2} \times \left( {70.0{\rm{ kg}}} \right) \times {\left( 0 \right)^2}\\ &= 3500{\rm{ J}}\end{aligned}\)

The power output of the sprinter is,

\(P = \frac{W}{t}\)

Here, t is the time required to attain the final velocity \(\left( {t = 3.00{\rm{ s}}} \right)\).

Putting all known values,

\(\begin{aligned}P& = \frac{{3500{\rm{ W}}}}{{3.00{\rm{ s}}}}\\ &= 1166.67{\rm{ W}}\end{aligned}\)

Converting power output of the sprinter from watts to horsepower,

\(\begin{aligned}P& = \left( {1166.67{\rm{ W}}} \right) \times \left( {\frac{{1{\rm{ hp}}}}{{746{\rm{ W}}}}} \right)\\ &= 1.56{\rm{ hp}}\end{aligned}\)

Therefore, the required power output of the sprinter is \(1166.67{\rm{ W}}\) or \(1.56{\rm{ hp}}\).

Comparing power output

(b)

No, a well-trained athlete cannot give this much power output for a longer time because a human is much less powerful. He cannot perform with so much power for longer period of time.

Most popular questions for Physics Textbooks

A toy gun uses a spring with a force constant of \(300{\rm{ N}}/{\rm{m}}\) to propel a \(10.0 - {\rm{g}}\) steel ball. If the spring is compressed \(7.00{\rm{ cm}}\) and friction is negligible:

(a) How much force is needed to compress the spring?

(b) To what maximum height can the ball be shot?

(c) At what angles above the horizontal may a child aim to hit a target \(3.00{\rm{ m}}\) away at the same height as the gun?

(d) What is the gun’s maximum range on level ground?

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

(c) What is the average force of friction if the hill has a slope 2.5° above the horizontal?

a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. b) What is the work done on the lift by the gravitational force in this process? c) What is the total work done on the lift?

Do devices with efficiencies of less than one violate the law of conservation of energy? Explain

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a 30° slope neglecting friction:

(a) Starting from rest.

(b) Starting with an initial speed of 2.50 m/s.

(c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.
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