Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
ass of 0.5 kilograms, falls off the edge of the scaffolding. Calculate the bucket’s kinetic and potential energy when it is 4 meters above the ground. Also calculate its velocity at this point.
2 answers:
Answer:
The bucket's velocity is
Explanation:
Given
Height above the ground, H = 12 m
Mass of bucket, m = 0.5 kg
Height of bucket, h = 4m
Taking acceleration of gravity, g as 9.8m/s²
First, we calculate the potential energy when bucket is on scaffolding (just about to fall) its total energy is 100%
P.E = mgH
P.E = 0.5 * 9.8 * 12
P.E = 58.8 J
At this point, the object is at rest, so the kinetic energy is 0 J
When the object is at 4 m above the ground, the bucket's total mechanical energy is as follows using conservation of energy
M.E
Where Total mechanical energy (M.E) = P.E (calculated above) = 58.8 J
By substitution, we have
--- Take Square roots
---- Approximated
Hence, the bucket's velocity is
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
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a) 893 N
b) 8.5 m/s
c) 3816 W
d) 69780 J
e) 8030 W
Explanation:
a)
The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:
where
m is Bolt's mass
a is the acceleration
In the first 0.890 s of motion, we have
m = 94.0 kg (Bolt's mass)
(acceleration)
So, the net force is
And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).
b)
Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
In the first phase of Bolt's race we have:
u = 0 m/s (he starts from rest)
(acceleration)
t = 0.890 s (duration of the first phase)
Solving for v,
c)
First of all, we can calculate the work done by Bolt to accelerate to a speed of
v = 8.5 m/s
According to the work-energy theorem, the work done is equal to the change in kinetic energy, so
where
m = 94.0 kg is Bolt's mass
v = 8.5 m/s is Bolt's final speed after the first phase
is the initial kinetic energy
So the work done is
The power expended is given by
where
t = 0.890 s is the time elapsed
Substituting,
d)
First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.
In the first 0.890 s, the force exerted was
We know that the average force for the whole race is
Which can be rewritten as
And solving for , we find the average force exerted by Bolt on the ground during the second phase:
The net force exerted by Bolt during the second phase can be written as
(1)
where D is the air drag.
The net force can also be rewritten as
where
is the acceleration in the second phase, with
u = 8.5 m/s is the initial speed
v = 12.4 m/s is the final speed
t = 8.69 t is the time elapsed
Substituting,
So we can now find the average drag force from (1):
So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:
where
d is Bolt's displacement in the second part, which can be found by using suvat equation:
And so,
e)
The power that Bolt must expend just to voercome the drag force is given by
where
is the increase in internal energy due to the air drag
t is the time elapsed
Here we have:
t = 8.69 s is the time elapsed
Substituting,
And we see that it is about twice larger than the power calculated in part c.