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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Answer:
Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]
Explanation:
To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.
The gravitational force is equal to:
Fg = (10 * 9.81) = 98.1 [N]
Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.
N - Fg = 0
N = Fg
N = 98.1 [N]
The friction force is defined as the product of normal force by the coefficient of friction.
Ff = N * μ
Ff = 98.1 * 0.4
Ff = 39.24 [N]
By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.
60 - Ff = m * a
60 - 39.24 = 10 * a
a = 2.076[m/^2]
The force exerted on the box is 56 N
Explanation:
The work done by a force on an object is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the box in this problem, we have:
W = 2240 J is the work done
d = 40 m is the displacement of the box
Assuming that the force is parallel to the displacement,
Solving the equation for F, we find the force exerted on the box:
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Answer:
None.
Explanation:
- Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
- As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:
- As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.