Answer:
Uā = 20 J
KEā = 40 J
v= 12.64 m/s
Explanation: Ā
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
Uā = m g H
Uā = 0.5 x 10 x 12 Ā Ā Ā Ā ( take g= 10 m/sĀ²)
Uā = 60 J
The potential energy at position 2
Uā = m g h
U ā= 0.5 x 10 x 4 Ā Ā Ā Ā ( take g= 10 m/sĀ²)
Uā = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m VĀ²
From energy conservation
Uā+KEā=Uā+KEā
By putting the values
60 - 20 = KEā
KEā = 40 J
lets take final velocity is v m/s
KEā= 1/2 m vĀ²
By putting the values
40 = 1/2 x 0.5 x vĀ²
160 = vĀ²
v= 12.64 m/s
Answer:
The bucket's velocity is [tex]v = 12.522 m/s[/tex]
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/sĀ²
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 [tex]= mgh + \frac{1}{2} mv^{2}[/tex]
Where Total mechanical energy (M.E) = P.E (calculated above) = 58.8 J
By substitution, we have
[tex]58.8 = 0.5 * 9.8 * 4 + \frac{1}{2} * 0.5 * v^{2}[/tex]
[tex]58.8 = 19.6 + \frac{1}{2} * 0.5 * v^{2}[/tex]
[tex]58.8 - 19.6 = \frac{1}{2} * 0.5 * v^{2}[/tex]
[tex]39.2 = \frac{1}{2} * 0.5 * v^{2}[/tex]
[tex]39.2 = \frac{1}{2} * \frac{1}{2} * v^{2}[/tex]
[tex]39.2 * 2 * 2= v^{2}[/tex]
[tex]156.8= v^{2}[/tex] --- Take Square roots
[tex]\sqrt{156.8} = \sqrt{v^{2}}[/tex]
[tex]v = \sqrt{156.8}[/tex]
[tex]v = 12.522 m/s[/tex] ---- Approximated
Hence, the bucket's velocity is [tex]v = 12.522 m/s[/tex]
