## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 6

#### Answer

The speed at the bottom of the incline is $6.0~m/s$

#### Work Step by Step

We can use conservation of energy to find the speed at the bottom of the incline: $KE_f +U_f= U_0+KE_0$ $\frac{1}{2}mv_f^2+0 = mgh+\frac{1}{2}mv_0^2$ $v_f^2 = 2gh+v_0^2$ $v_f = \sqrt{2gh+v_0^2}$ $v_f = \sqrt{(2)(9.80~m/s^2)(2.0~m)~sin~30.0^{\circ}+(4.0~m/s)^2}$ $v_f = 6.0~m/s$ The speed at the bottom of the incline is $6.0~m/s$.

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