Question

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.180 m from their uncompressed length.

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

Answer 1

To hold the platform in position, you must apply a force of approximately 888.89 N, calculated using the work done to compress the springs and Hooke's law.

Step-by-step explanation:

To calculate the magnitude of force you must apply to hold the springs in a compressed position, we can use the work done on the springs. Given that you do 80.0 J of work to compress the springs by 0.180 m, we can find the force using the formula for work done on a spring:

W = ½ k x²

Where W is the work done, k is the spring constant, and x is the displacement from the spring's equilibrium position. We rearrange the formula to solve for the force constant (k):

k = 2W / x²

Plugging in the values we have:

k = 2(80.0 J) / (0.180 m)²

k = 160.0 J / 0.0324 m²

k = 4938.27 N/m

Now that we have the spring constant, we can use Hooke's law (F = -kx) to find the force needed to hold the springs compressed:

F = kx

F = (4938.27 N/m)(0.180 m)

F = 888.89 N

Thus, you must apply a force of approximately 888.89 N to hold the platform in position.

Answer 2

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Step-by-step explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation for this work done by the spring is expressed as:

`W = (1)/(2)k_(eq)x^2`

Making the spring constant
`k_(eq)` the subject of the formula; we have:

`k_(eq) = (2W)/(x^2)`

Substituting our given values, we have:

`k_(eq) = (2*80)/(0.180^2)`

`k_(eq) = 4938.27 \ N.m^(-1)`

The magnitude of the force that must be apply to the hold platform in this position is given by the formula :

`F = k_(eq)x`

`F = 4938.27*0.180`

F = 888.89 N