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 83.0 J when you compress the springs a distance of 0.230 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 0.230 m farther? What maximum force must you apply to move the platform a distance 0.230 m farther?

Answer 1

Answer:

a) $F_{person} = 360.87\,N$, b) $W_{person} = 332\,J$, c) $F_{person} = 721.739\,N$

Explanation:

a) The system can be described by the Principle of Energy Conservation and Work-Energy Theorem:

$U_{A} + W_{person} = U_{B}$

The work done to compress the springs is:

$W_{person} = 83\,J$

The force needed to compress the springs is:

$F_{person} = \frac{83\,J}{0.23\,m}$

$F_{person} = 360.87\,N$

b) Spring potential energy is directly proportional to the squared of elongation. Then:

$U_{B} - U_{A} = 83\,J\cdot \frac{(0.46\,m)^{2}}{(0.23\,m)^{2}}$

$U_{B} - U_{A} = 332\,J$

$W_{person} = 332\,J$

c) The force needed to compress the springs is:

$F_{person} = \frac{332\,J}{0.46\,m}$

$F_{person} = 721.739\,N$