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gymnast of mass 51.00 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.100 m above the trampoline and, when she lands, her feet stretch the trampoline down 61.00 cm. How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]

Respuesta :

Answer:

The displacement is 0.1452 m.

Explanation:

Given that,

Mass of gymnast = 51.00 kg

Maximum height = 2.100 m

Minimum displacement = 61.00 cm

We need to calculate the displacement

Using restoring force

[tex]F = kx[/tex]

[tex]mg=kx[/tex]

[tex]x=\dfrac{mg}{k}[/tex]....(I)

Now using gravitational potential energy

[tex]U_{g}=mgh[/tex]

[tex]\dfrac{1}{2}kx_{min}^2=mgh[/tex]

[tex]k=\dfrac{2mgh}{x_{min}^2}[/tex]

Put the value of k in equation (I)

[tex]x=\dfrac{mg}{\dfrac{2mgh}{x_{min}^2}}[/tex]

[tex]x=\dfrac{x_{min}^2}{2h}[/tex]

Put the value into the formula

[tex]x=\dfrac{61.00\times10^{-2}}{2\times2.100}[/tex]

[tex]x=0.1452\ m[/tex]

Hence, The displacement is 0.1452 m.

onyebuchinnaji

The stretch of the trampoline when she stands on it at rest is 8.8 cm.

The given parameters;

  • mass of the gymnast, m = 51 kg
  • height reached by the gymnast, h = 2.1 m
  • extension of the spring, x = 61 cm = 0.61 m

Apply the principle of conservation of energy;

[tex]mgh = \frac{1}{2}kx^2\\\\2mgh = kx^2\\\\k = \frac{2mgh}{x^2} \ ---(1)[/tex]

From Newton's second law of motion and Hook's law;

mg = kx

mg = kd

[tex]k = \frac{mg}{d} \ ---(2)[/tex]

where;

  • d is the stretch of the trampoline when she stands on it at rest;

solve (1) and (2) together;

[tex]\frac{mg}{d} = \frac{2mgh}{x^2} \\\\2mghd = mgx^2\\\\2hd = x^2\\\\d = \frac{x^2}{2h} \\\\d = \frac{(0.61)^2}{2 \times 2.1} \\\\d = 0.088 \ m[/tex]

d = 8.8 cm

Thus, the stretch of the trampoline when she stands on it at rest is 8.8 cm.

Learn more here:https://brainly.com/question/4404276

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