A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this high? (The acceleration due to gravity is 9.8 m/s2.) Show your work.
+33616 From conservation of energy, the kinetic energy at take off equals the potential energy at the top of the jump, which is just mgh: 7.2*9.8*1.2 Joules
$${\mathtt{Energy}} = {\mathtt{7.2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{1.2}} \Rightarrow {\mathtt{Energy}} = {\mathtt{84.672}}$$ Joules
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+33616 From conservation of energy, the kinetic energy at take off equals the potential energy at the top of the jump, which is just mgh: 7.2*9.8*1.2 Joules
$${\mathtt{Energy}} = {\mathtt{7.2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{1.2}} \Rightarrow {\mathtt{Energy}} = {\mathtt{84.672}}$$ Joules
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