Physics, 21.02.2020 02:33, gabby1274

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 81.0 J when you compress the springs a distance of 0.250 m from their uncompressed length.

Part A

What magnitude of force must you apply to hold the platform in this position?

Part B

How much additional work must you do to move the platform a distance 0.250 m farther?

Part C

What maximum force must you apply to move the platform to the position in Part B?

Answers: 2

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Answers

answered: u8p4

1. Part A: 648N2. Part B: 324J3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

answered: Guest

The horse latitudes-high-pressure zones about 30 degrees north and 30 degrees south of the equator.-warm air traveling away from the equator cools and sinks-weather tends to be clear and drythe doldrums-low-pressure near equator-air rises and spreads out toward the poles-rising moist air produces clouds and heavy rainthe westerlies-blow from the west-move from the horse latitudes toward the poles.-bring storms across much of the united statesthe tradewinds-blow from the east-move from the horse latitudes toward the equator -strong, steady winds

answered: Guest

cell differentiation is how generic embryonic cells become specialized cells. this occurs through a process called gene expression. gene expression is the specific combination of genes that are turned on or off (expressed or repressed), and this is what dictates how a cell functions.

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As part of your daily workout, you lie on your back and push with your feet against a platform attac...