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
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
10 answers
for part c), I made a typo:
KE = (1/2)*(15)*(12.53)^2 = 1177.2J
KE = (1/2)*(15)*(12.53)^2 = 1177.2J
where did u get 12.53m/s
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mga kumukuha ng sagot diyan mga grade 11 hule kayoo
grade 12 pala nakopo hulee
ang gulo ng explanation pero okay lang haha
3 answers