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goldfiish [28.3K]
2 years ago
11

A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is

Physics
1 answer:
prisoha [69]2 years ago
5 0

0J

Explanation:

A wall that is acted upon by a force of 20N does not move suggests that no work has been done on the wall.

Work done is the force applied to move in the direction of the applied force.

   Work done = force x distance.

Since the distance is 0, the work done on the body too is zero.

Work is only done when forces moves a body through a given distance.

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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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
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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

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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

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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

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