Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 168: 48


a) $H=2.5m$ b) $H_1+H_2=1.98m$

Work Step by Step

a) As the block slides upward along the longer track, its KE decreases while PE increases. From the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mgH$$ $$\frac{1}{2} (v_f^2-v_0^2)=-gH (1)$$ The block has initial speed $v_0=7m/s$ and final speed $v_f=0$. From here, we can find $H$. $$\frac{1}{2} (0-7^2)=-9.8H$$ $$H=2.5m$$ b) First, we find the speed at which the block leaves the track $v_f$, using equation (1): $$\frac{1}{2} (v_f^2-v_0^2)=-gH_1$$ We have $v_0=7m/s$ and $H_1=1.25m$, so $$\frac{1}{2} (v_f^2-7^2)=-9.8\times1.25=-12.25$$ $$v_f=4.95m/s$$ The block leaves the track at angle $\theta=50^o$, so its vertical speed is $v_y=4.95\sin50=3.79m/s$. If we consider only vertical movement, the block reaches its highest point when $v_y=0$. So we can take $v_0=3.79m/s$, $v_f=0$ and use equation (1) to find $H_2$ $$\frac{1}{2} (0-3.79^2)=-9.8H_2$$ $$H_2=0.73m$$ So, $H_1+H_2=1.98m$
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