A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
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Answer:
1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>
Explanation:
We note that the total mechanical energy (M.E.) of the body is given as follows;
M.E. = K.E. + P.E. = Constant
Where;
K.E. = The kinetic energy of the body = (1/2)·m·v²
P.E. = The potential energy of the body = m·g·h
m = The mass of the person
v = The velocity with which the person is in motion
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height of the person above the ground
The length of the rope = 20 m
The initial height at location 1, h₁ = 40.0 m
At location 1, the velocity, v₁ = 0.00 m/s
The mechanical energy, M.E. = K.E.₁ + P.E.₁
∴ K.E.₁ = 0 and P.E.₁ = m ×9.81×40
M.E. = (1/2) ×m ×0² + m ×9.81×40
∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy
At location 2, the velocity, v₂ = 10.0 m/s
The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40
∴ K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m
M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.
At location 3, the velocity, v₃ = 20.0 m/s
The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20
∴ K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m
M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.
At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m
The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15
∴ K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m
M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.
At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m
The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10
∴ K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m
M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.
Therefore, we have;
Average speed is defined as total distance moved in total interval of time
so it is given as
now here is we show distance by "d" and time by"t"
then we will have mathematical expression as follows