Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.6 - Conservation of Energy - Problems - Page 226: 77

$h=23.75m$, $v_C=21.6m/s$

Work Step by Step

We can determine the required height and speed as follows: We apply the equation of conservation of energy between A and B $\frac{1}{2}m_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B$ We plug in the known values to obtain: $0+9.81h=\frac{1}{2}(8.578)^2+(9.81)(20)$ This simplifies to: $h=23.75m$ Now, we apply the equation of conservation of energy between B and C to determine the speed $\frac{1}{2}mv_B^2+mgh_B=\frac{1}{2}mv_C^2+mgh_C$ We plug in the known value to obtain: $\frac{1}{2}(8.578)^2+(9.81)(20)=\frac{1}{2}v_C^2+0$ This simplifies to: $v_C=21.6m/s$

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