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When a 61 kg cheerleader stands on a vertical spring, the spring compresses by 5.8 cm. when a second cheerleader stands on the shoulders of the first, the spring compresses an additional 4.2 cm?

Respuesta :

Let m kg be the mass of the second cheerleader.

By definition, the spring constant is
[tex]k = \frac{(61 \, kg)(9.8 \, m/s^{2})}{0.058 \, m} = \frac{(61+m \, kg)(9.8 \, m/s^{2})}{0.058+0.042 \, m} [/tex]

Therefore
m + 61 = (0.1)*(1.0517 x 10³) = 105.172
m = 44.172 kg

Answer: 44.2 kg

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