Asked by k

a 15kg rock falls from the top of a building 8.0 high. Find its PE and KE
a.when at rest at the top of the building
b. halfway down the ground
c. when it reaches the ground
d.speed when it reaches the ground
Answered by DonHo
a) when at rest
PE=mgh
h=8m
mgh = (15kg)(9.81m/s^2)(8m)=1177.2 J

KE = (1/2)*m*v^2
there is no velocity, v=0
so KE = 0J

b)
PE=(15kg)*(9.81m/s^2)(8/2m) = ?J
There is no KE while it's falling
KE = 0J

c) PE= m*g*h = (15)(9.81)(0) =0J

KE =(1/2)m*v^2
where v=sqrt(2*g*h)
= (2*9.81*8) = 12.53 m/s

KE = (1/2)*(15)*(156.96)^2 = ? J

d) speed when it reaches the ground:
v=12.53 m/s


To check:

PE_initial + KE_initial = PE_final + KE_final



1177.2J + 0J = 0J + 1177.2J
Answered by DonHo
for part c), I made a typo:

KE = (1/2)*(15)*(12.53)^2 = 1177.2J
Answered by k
where did u get 12.53m/s
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