Question

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 82.0 J when you compress the springs a distance of 0.220 m from their uncompressed length. What magnitude of force must you apply to hold the platform in this position? How much additional work must you do to move the platform a distance 0.220 m farther? What maximum force must you apply to move the platform a distance 0.220 m farther?

Answer 1

Answer:

F =1490.9 N

Explanation:

given,

work done = 82 J

compression length = 0.220 m

for parallel combination

$k_{eq}= k_1+k_2$

         = 2 k

work done

$w = \dfrac{1}{2}k_{eq}x^2$

$k_{eq} = \dfrac{2W}{x^2}$

$k_{eq} = \dfrac{2\times 82}{0.22^2}$

$k_{eq} =3388.43 N/m$

force =

F = kx

F = 3388.43 ×0.22

F =745.45 N

the additional work done

final potential energy = $\dfrac{1}{2}k_{eq}(2x)^2$

                                    = 4 W = 328 J

the additional work = 328 - 82 = 255 J

maximum force

F = k × 2x

F = 2×745.45

F =1490.9 N

                                   

Answer 2

The reaction of the spring gives the amount of force that must

be applied to hold the spring in position.

Responses (approximate values);

• The force to be applied is 745.45 N

• Additional work is 246.0 J

• The maximum force is 1490.9 N

Which methods are used to determine the force in the spring?

First question:

The amount of work done to compress the spring to 0.220 m = 82.0 J

The magnitude of the force that must be applied to hold the

the platform at 0.220 m from the uncompressed length is

given as follows;

$Energy \ in \ spring = \mathbf{\dfrac{1}{2} \cdot k \cdot x^2}$

Where;

x = Extension of the spring

By the principle of conservation of energy, therefore;

$82.0 = \mathbf{\dfrac{1}{2} \times k \times 0.22^2}$

Which gives;

k ≈ 3388.43 N/m

According to Hooke's law;

Force in spring = k × x

According to Newton's third law of motion

Force required = Force in spring

Therefore;

Force required ≈ 3388.43 N/m × 0.220 m N = 745.45 N

• The force that must be applied, F ≈ 745.45 N

Second question

The distance moved 0.220 m farther = 0.220 m + 0.220 m = 0.440 m

$Work \ required = \mathbf{\dfrac{1}{2} \times 3388.43 \times 0.44^2} = 328$

Additional work = Total work - Initial work

Therefore;

• Additional work = 328.0 J - 82.0 J = 246.0 J

Third question

Force applied = Force in spring

Which gives;

$Force applied = \dfrac{1}{2} \times 3388.43 \times 0.44 \approx \mathbf{1490.9}$

• The force that must be applied, F ≈ 1,490.9 N

Learn more about Hooke's law here;

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