Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 276: 34


The string stretches by $1.945m$.

Work Step by Step

1) The climber's velocity after he falls $0.75m$ is $$v_f^2=v_0^2+2gs$$ $$v_f^2=0+2(9.8)(0.75)=14.7$$ $$v_f=3.83m/s$$ 2) Because energy is conserved, $$E_0=E_f (1)$$ We take the initial point to be when the rope runs out of slack, so $h_0=0$ and $y_0=0$. The climber's speed $v_0=3.83m/s$ The final point is when the rope stretches to the maximum of length $y_f (y_f\lt0$ because the climber falls). The climber's height then $h_f=y_f$. The climber's speed $v_f=0$ From (1), we have: $$mgy_f+\frac{1}{2}ky_f^2=\frac{1}{2}mv_0^2$$ The climber's weight $mg=842.8N$ and $k=1.2\times10^3N/m$. The initial KE $\frac{1}{2}mv_0^2=\frac{1}{2}(86)(3.83)^2=630.76J$. Therefore, $$(0.6\times10^3)y_f^2+842.8y_f-630.76=0$$ After solving the equation, because $y_f\lt0$, we get $y_f=-1.945m$
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