Question

In a volcanic eruption, a 2.0 kg piece of porous volcanic rock is thrown vertically upward with an initial speed of 44 m/s. It travels upward a distance of 47 m before it begins to fall back to the Earth.

1) (a) What is the initial kinetic energy of the rock? (in J)

2) (b) What is the increase in thermal energy due to air friction during ascent? (in J)

3) (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? (in m/s)

Answer 1

1(a) Ki = 0.5*m*vi^2

= 0.5*2*44^2

= 1936 J <<<<<<---------Answer

2(b) Increase in thermal enrgy = Ki - m*g*h

= 1936 - 2*9.8*47

= 1014.8 J <<<<<<---------Answer

3(c) Kf = Ki - (1014.8 + 0.7*1014.8)

= 1936 - (1014.8 + 0.7*1014.8)

= 210.84 J

Apply, Kf = 0.5*m*vf^2

==> vf = sqrt(2*Kf/m)

= sqrt(2*210.84/2)

= 14.5 m/s <<<<<<---------Answer

Answer 2

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?

Reference https://www.physicsforums.com/threads/kinetic-thermal-energy-problem.262442/