In a volcanic eruption, a 2.0 kg piece of porous volcanic rock is thrown vertica

Question

In a volcanic eruption, a 2.0 kg piece of porous volcanic rock is thrown vertically upward with an initial speed of 36 m/s. It travels upward a distance of 50 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?

Explanation / Answer

given m = 2kg

vi = 36 m/s

d = 50 m

a. initial KE of the rock = 0.5mvi^2 = 1296 J

b. increase in thermal energy = E

form energy balance

1296 - E = m*g*d

E = 315 J

c. energy decrease while coming down

E' = 0.7E

then final speed = u

0.5mu^2 = mgd - E'

u = 27.5771644 m/s

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