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A 4kg block is sitting on the floor.
(a) How much potential energy does it have? (b) How much kinetic energy does it have?
(c) The block is raised to 2m high. How much potential energy does it have? (d) The block is raised to 4m high. How much potential energy does it have?
(e) The block is raised to 45m high. How much potential energy does it have? (f) The block is dropped to the ground. How fast is it traveling when it hits the ground? (remember chapter 2?)
(g) How much kinetic energy does it have when it hits the ground?

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(a)
recall that potential energy is stored energy, and is given by the formula:
PE = mgh (units in Joules)
where
m = mass (in kg)
g = acceleration due to gravity = 9.8 m/s^2
h = height (in m)
if the reference position is at the floor (that is, h = 0), the PE is equal to
PE = 4*9.8*0
PE = 0

(b)
recall that kinetic energy is energy in motion and is given by the formula:
KE = (1/2)*m*v^2
where
v = velocity (in m/s)
since it's not moving (v = 0),
KE = 0

(c)
at 2 m high,
PE = 4*9.8*2
PE = 78.4 J

(d)
at 4 m high,
PE = 4*9.8*4
PE = 158.8 J

(e)
at 45 m high,
PE = 4*9.8*45
PE = 1764 J

(f)
recall that the motion of the block is uniformly accelerated motion, and we can therefore use the formula:
(v,f)^2 - (v,o)^2 = 2gh
where
v,f = final velocity (in m/s)
v,o = initial velocity (in m/s)
since it is dropped from rest, v,o = 0:
v,f^2 = 2gh
v,f = sqrt(2gh)
v,f = sqrt(2*9.8*45)
v,f = 29.7 m/s

(g)
KE = (1/2)mv^2
KE = (1/2)*4*29.7^2
KE = 1764.18 J

this actually shows the law conservation of energy, which is ΔPE = -ΔKE

hope this helps~ :)

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