# Kinetic/Thermal Energy Problem

by maniacp08
Tags: energy, kinetic or thermal
 P: 116 In a volcanic eruption, a 2.0 kg piece of porous volcanic rock is thrown vertically upward with an initial speed of 45 m/s. It travels upward a distance of 45 m before it begins to fall back to the Earth. (a) What is the initial kinetic energy of the rock? (b) What is the increase in thermal energy due to air friction during ascent? (c) If the increase in thermal energy due to air friction on the way down is 70% of that on the way up, what is the speed of the rock when it returns to its initial position? Relevant Equations: K = 1/2 M * V^2 External Work = Change in Mechanical Energy + Change in Thermal Energy Change in Thermal Energy = Friction Force * Displacement Is the answer for part A just 1/2 M * V^2? It seems a little bit too easy. For Part B- Is to find the change in thermal energy during its ascent Change in Thermal Energy = Friction Force * Displacement Friction Force = Uk * Normal Force How would I find Uk or Normal Force? Any help would be grateful.
Mentor
P: 41,489
 Quote by maniacp08 Is the answer for part A just 1/2 M * V^2? It seems a little bit too easy.
They can't all be hard.

 For Part B- Is to find the change in thermal energy during its ascent Change in Thermal Energy = Friction Force * Displacement Friction Force = Uk * Normal Force How would I find Uk or Normal Force?
Kinetic friction is a model that applies to dry solids rubbing against each other; it doesn't apply here. Luckily, you don't need it. Consider the total mechanical energy the rock had to start with (part A) and compare it to what's left when it reaches its highest point.
 P: 116 If I let y=0 when the rock is launched Ki + Ui = Kf + Uf? The initial Kinetic energy is 1/2 * M * Vi^2 The initial Potential Energy is 0 Is the Final Kinetic Energy where it reaches its highest point 0 since the final velocity will be 0 at its highest point The final potential energy is MGH where H is 45m So I have 1/2 M * Vi^2 = MGH What am I suppose to solve/find out?
P: 1,970
Kinetic/Thermal Energy Problem

 Quote by maniacp08 So I have 1/2 M * Vi^2 = MGH What am I suppose to solve/find out?
Nothing, you have everything. Did you check if it's true? I mean, are the two sides equal???
I don't think so. Why?
 P: 116 Ahh, I got it, the difference is 1142.1 J. Thanks! For Part C: (c) If the increase in thermal energy due to air friction on the way down is 70% of that on the way up, what is the speed of the rock when it returns to its initial position? 70% of 1142.1J = 799.47J Would it be Ki + Ui = Kf + Uf From the highest point to the initial position? 0 + MGH = 1/2 M * Vf^2 + 0 MGH = 1/2 M * Vf^2 MGH + 799.47 = 1/2 M * Vf^2 and solve for Vf?
 Mentor P: 41,489 Almost, but you're getting a bit mixed up. Think of the energy this way: What you start with (PE at the top) = KE at the bottom (1/2MVf^2) + Thermal energy (which you've just calculated) Rearrange that and solve for Vf.
 P: 116 Thanks Doc!

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