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College Physics (Urone)
Found in: Page 263
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

Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs.

(a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.)

(b) Compare this force with the weight of the jogger.

(a) The magnitude of the force needed to stop the downward motion of a jogger’s leg is \(16335{\rm{ N}}\).

(b) The ratio of the total force to the weight of the jogger is \(22.2\) times.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Weight

Weight is a force which acts on a mass due to acceleration due to gravity.

Mathematically,

\(W = mg\)

Here, m is the mass of the object, and g is the acceleration due to gravity.

Step 2: Magnitude of the force

(a)

The force on leg is,

\(\begin{aligned}{c}F = {m_l}a\\ = \frac{{{m_l}{v^2}}}{{2s}}\end{aligned}\)

Here, \({m_l}\) is the mass of the leg \(\left( {{m_l} = 13.0{\rm{ kg}}} \right)\), v is the velocity \(\left( {v = 6.00{\rm{ m}}/{\rm{s}}} \right)\) and s is the stopping distant \(\left( {s = 1.50{\rm{ cm}}} \right)\).

Putting all known values,

\(\begin{aligned} F &= \frac{{\left( {13.0{\rm{ kg}}} \right) \times {{\left( {6.00{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( {1.50{\rm{ cm}}} \right)}}\\ &= \frac{{\left( {13.0{\rm{ kg}}} \right) \times {{\left( {6.00{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( {1.50{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}\\ &= 15600{\rm{ N}}\end{aligned}\)

The weight of the body is,

\({W_g} = {m_b}g\)

Here, \({m_b}\) is the mass of the body \(\left( {{m_b} = 75{\rm{ kg}}} \right)\), and g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\).

Putting all known values,

\(\begin{aligned} {W_g} &= \left( {75.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ &= 735{\rm{ N}}\end{aligned}\)

The total force on the body is,

\({F_{total}} = F + {W_g}\)

Putting all known values,

\(\begin{aligned} {F_{total}} &= \left( {15600{\rm{ N}}} \right) + \left( {735{\rm{ N}}} \right)\\ &= 16335{\rm{ N}}\end{aligned}\)

Therefore, the magnitude of the force needed to stop the downward motion of a jogger’s leg is \(16335{\rm{ N}}\).

Step 3: Compare the ratio of the force to the weight

(b)

The ratio of the total force to the weight is,

\(\begin{aligned} \frac{{{F_{total}}}}{{{W_g}}} &= \frac{{16335{\rm{ N}}}}{{735{\rm{ N}}}}\\ &= 22.2\end{aligned}\)

Therefore, the ratio of the total force to the weight is \(22.2\).

Most popular questions for Physics Textbooks

Question: Mountain climbers carry bottled oxygen when at very high altitudes.

(a) Assuming that a mountain climber uses oxygen at twice the rate for climbing 116 stairs per minute (because of low air temperature and winds), calculate how many liters of oxygen a climber would need for 10.0 h of climbing. (These are liters at sea level.) Note that only 40% of the inhaled oxygen is utilized; the rest is exhaled.

(b) How much useful work does the climber do if he and his equipment have a mass of 90.0 kg and he gains 1000 m of altitude?

(c) What is his efficiency for the 10.0-h climb?

Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less than needed to fuel metabolism. The manufacturers of an exercise bicycle claim that you can lose 0.500 kg of fat per day by vigorously exercising for 2.00 h per day on their machine.

(a) How many kcal are supplied by the metabolization of 0.500 kg of fat?

(b) Calculate the kcal/ min that you would have to utilize to metabolize fat at the rate of 0.500 kg in 2.00 h.

(c) What is unreasonable about the results?

(d) Which premise is unreasonable, or which premises are inconsistent?

(a) Calculate the force the woman in Figure 7.46 exerts to do a push-up at constant speed, taking all data to be known to three digits.

(b) How much work does she do if her center of mass rises 0.240 m?

(c) What is her useful power output if she does 25 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.

Figure 7.46 Forces involved in doing push-ups. The woman’s weight acts as a force exerted downward on her center of gravity (CG).

The swimmer shown in Figure 7.44 exerts an average horizontal backward force of 80.0 N with his arm during each 1.80 m long stroke.

(a) What is his work output in each stroke?

(b) Calculate the power output of his arms if he does 120 strokes per minute.

Figure 7.44

(a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months?

(b) How long can a battery that can supply 8.00×104 J run a pocket calculator that consumes energy at the rate of 1.00×10−3 W?
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