You throw a 4kg rock directly down at 6m/s from the top of a 18m cliff. Calculat

Question Description

I don’t know how to handle this Physics question and need guidance.

a)initial KE

b)Initial PE

c) V when rock is 13m above ground

d) V rock will hit the ground 

e)distance the rock is driven into the mud if the mud exerts a steady upwards force of 4000N 

f)speed the rock hits ground if initially thrown up at 6m/s

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Final Answer

a) Potential energy is mgh where h = vertical distance. 
b) Kinetic energy is (1/2)mv^2 
c) It has gained energy mgh where h is the vertical distance it fell (so far). Add that to the initial KE. 
d) It has gained energy mgh where h is the total vertical distance it fell. Add that to the initial KE. 
e) Work F * d = kinetic energy, because the force has to remove that kinetic energy from the rock. 
f) Exactly the same

madhavant (2757)
UC Berkeley

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