# Chapter 10 - Interactions and Potential Energy - Exercises and Problems - Page 258: 41

(a) The speed at the top is 2.3 m/s (b) The ice cube's speed at an angle of $\theta$ is $\sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$

#### Work Step by Step

(a) We can use conservation of energy to solve this question. Note that the height at the top of the pipe is 20 cm. $KE_2+PE_2 = KE_1+PE_1$ $\frac{1}{2}mv_2^2+mgh=\frac{1}{2}mv_1^2+0$ $v_2^2=v_1^2-2gh$ $v_2= \sqrt{v_1^2-2gh}$ $v_2= \sqrt{(3.0~m/s)^2-(2)(9.80~m/s^2)(0.20~m)}$ $v_2 = 2.3~m/s$ The speed at the top is 2.3 m/s (b) We can find the height $h$ above the bottom of the pipe when the angle is $\theta$. Let $R$ be the radius of the pipe. $\frac{R-h}{R} = cos(\theta)$ $h = R~(1-cos(\theta))$ We can use conservation of energy to find the speed when the ice cube is at an angle $\theta$. Note in part (a) that: $v_2= \sqrt{v_1^2-2gh}$ $v_2= \sqrt{v_1^2-2gR~(1-cos(\theta))}$ $v_2= \sqrt{v_1^2- 2gR+2gR~cos(\theta)}$ $v_2= \sqrt{(3.0~m/s)^2-(2)(9.80~m/s^2)(0.10~m)+(2)(9.80~m/s^2)(0.10~m)~cos(\theta)}$ $v_2= \sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$ The ice cube's speed at an angle of $\theta$ is $\sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.