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Maurinko [17]
2 years ago
14

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

is 10% efficient at converting chemical energy in your body into mechanical work.
A.) What is your internal chemical energy change?

B.) How long must you bike to convert 3.8×105 J of chemical potential while staying at this speed? (This amount of energy equals the energy released by the body after eating three slices of bread.)
Physics
1 answer:
stellarik [79]2 years ago
3 0

A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

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Answer:

Explanation:

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And the other 7M

And this a velocity V

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Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

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(M+7M) × 0 = 7M •V − Mv

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Divide both sides by M

0 = 7V -v

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Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

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Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

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Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

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