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A gymnast of mass 52.0 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.98 m above the trampoline and, when she lands, her feet stretch the trampoline 70.0 cm down. How far does the trampoline stretch when she stands on it at rest? Assume that the trampoline is described by Hooke’s law when it is stretched.

Sagot :

When the gymnast is at 2.98 meters above the trampoline, she has potential gravitational energy in relation to the trampoline:

[tex]\begin{gathered} PE_g=m\cdot g\cdot h \\ PE_g=52\cdot9.81\cdot2.98 \\ PE_g=1520.16\text{ J} \end{gathered}[/tex]

When she lands on the trampoline, all this energy will be converted into potential elastic energy:

[tex]\begin{gathered} PE_e=\frac{kx^2}{2} \\ 1520.16=\frac{k\cdot0.7^2}{2} \\ 0.49k=3040.32 \\ k=6204.73\text{ N/m} \end{gathered}[/tex]

Now, to find the stretch in the trampoline when the gymnast is at rest, let's use the gymnast weight force in the formula for the force in a string:

[tex]\begin{gathered} F=k\cdot x \\ m\cdot g=k\cdot x \\ 52\cdot9.81=6204.73\cdot x \\ 510.12=6204.73 \\ x=\frac{510.12}{6204.73} \\ x=0.0822\text{ m} \end{gathered}[/tex]

Therefore the trampoline stretches 8.22 cm.

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