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As part of your work out, 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 80.0 J of work when you compress the springs 0.200m from their uncompressed length. What magnitude of force must you apply to hold the platform in this position?

The aim of this question is to develop an understanding of the basic concepts of work done and resultant force.

The work done is a scalar quantity defined as the amount of energy dispensed whenever a forcing agent moves a body along some distance in the direction of the force. Mathematically, it’s defined as the dot product of force and displacement.

\[ W \ = \ \vec{ F } . \ \vec{ d } \]

Where W is the work done, F is the average force and d is the displacement. If force and displacement are colinear, then the above equation reduces to:

\[ W \ = \ | \vec{ F } | \times | \vec{ d } | \]

Where $ | \vec{ F } | $ and $ | \vec{ d } | $ are the magnitudes of force and displacement.

Whenever two or more forces act on a body, the body moves in the direction of the net force or resultant force. Net force or resultant force is the vector sum of all the forces acting on the said body. The net force can be calculated using vector addition methods such as a head-to-tail rule or polar coordinate addition or complex addition etc.

Expert Answer

Given that:

\[ \text{ Work Done } = \ W \ = \ 80 \ J \]

\[ \text{ Distance Covered } = \ d \ = \ 0.2 \ m \]

From the definition of work done, we can find the average force on one spring during this movement by utilizing the following formula:

\[ \text{ Work Done } = \text{ Average Force } \times \text{ Distance Covered } \]

\[ W \ = \ F \times \ d \]

\[ \Rightarrow F \ = \ \dfrac{ W }{ d } \ … \ …\ … \ ( 1 ) \]

Substituting given values:

\[ F \ = \ \dfrac{ 80 \ J }{ 0.2 \ m } \]

\[ \Rightarrow F \ = \ 400 \ N \]

Since there are two springs, so the net force needed to press both the springs by a distance of 0.2 m will be twice:

\[ F_{ net } \ = \ 2 \times 400 \ N \]

\[ \Rightarrow F_{ net } \ = \ 800 \ N \]

Numerical Result

\[ F_{ net } \ = \ 800 \ N \]

Example

Given the same platform, how much force will be needed to push the platform by a distance of 0.400 m from the uncompressed position?

Recall equation (1):

\[ \Rightarrow F \ = \ \dfrac{ W }{ d } \]

Substituting given values:

\[ F \ = \ \dfrac{ 80 \ J }{ 0.4 \ m } \]

\[ \Rightarrow F \ = \ 200 \ N \]

Since there are two springs, so the net force needed to press both the springs by a distance of 0.4 m will be twice:

\[ F_{ net } \ = \ 2 \times 200 \ N \]

\[ \Rightarrow F_{ net } \ = \ 400 \ N \]

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