# Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 11

The car's speed at the gas station will be 1.4 m/s.

#### Work Step by Step

We can use conservation of energy to find the speed $v_2$ at the gas station. Therefore; $KE_2+PE_2 = KE_1+PE_1$ $\frac{1}{2}mv_2^2+mg~h_2=\frac{1}{2}mv_1^2+mg~h_1$ $v_2^2=v_1^2+2g~(h_1-h_2)$ $v_2= \sqrt{v_1^2+2g~(h_1-h_2)}$ $v_2= \sqrt{(10~m/s)^2+(2)(9.80~m/s^2)(10~m-15~m)}$ $v_2 = 1.4~m/s$ The car's speed at the gas station will be 1.4 m/s.

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