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
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
where did u get 12.53m/s
Taena niyo ayos para sa activity ko to sa school eh
for part c), I made a typo:
KE = (1/2)*(15)*(12.53)^2 = 1177.2J