Pls explain in DETAIL. How did you come up with this solution. Please write a detailed explanation because I'm going to present it in class. I don't know about the terms pls
Pls explain in DETAIL. How did you come up with this solution. Please write a detailed explanation because I'm going to present it in class. I don't know about the terms pls
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Pls explain in DETAIL. How did you come up with this solution. Please write a detailed explanation because I'm going to present it in class. I don't know about the terms pls
Transcribed Image Text:1.
Given :
Height of bridge = 90 ft
1 ft = 0.3048 m
90 ft = (0.3048 × 90) m
90 ft = 27.432 m
1 ft = 0.3048 m90 ft = 0.3048 x 90 m90 ft = 27.432 m
Mass of player 101 = 83.2 kg
Mass of player 212 = 51.7 kg
Combined mass of player 101 and player 211 = 83.2 kg + 51.7 kg = 134.9 kg
(a) The speed of the two players as they hit the ground :
![The formula of mechanical energy :
М.Е 3D К. Е + Р.E
M.E = mv? + mgh
Where,
m = mass of an object
v = velocity of that object
g = acceleration due to gravity
h = height of an object.
Conservation of Mechanical Energy :
The total mechanical energy of a system is conserved
i. e. the energy can neither be created nor be destroyed.
So,
Mechanical energy at the bridge = Mechanical energy at the ground
M.Epridge = M.Eground
mcombined (v² bridge) + mcombined × g × (hbridge) = ; mcombinedv ground
+ mcombined X g Xx (hground)
x 134.9 x 0 + 134.9 × 9. 8 x 27.432 = x 134.9 x v² + 134.9 × 9. 8
134.9 x 9. 8 x 27. 432 = × 134. 9 × v²
v = 2x134.9x9.8x27432
134.9
V ground = V
2x134.9x9 8x27.432 mls
134.9
Vgroand = 23. 1877 m/s
(b) The total work done by gravity on the two players :
Total work done by gravity on the two players = Loss in Potential energy = Gain in Kinetic energy
AP.E = [mcombined × g × (hbridge)] – [mcombined X g × (hground)]
A P.E = [134.9 × 9. 8 × 27. 432 ] – [134.9 × 9. 8 × 0]
AP.E = 36265, 8 J
AK.E = [mcombined (Vground)] – [¿mcombined (Vbridge )]
A K. E = [ x 134. 9 × 23. 1877² ] – [ x 134. 9 x 0² ]
AK.E = 36265, 8 J
So Work Done by gravity = + 36265, 8 J
(c) The value of final Kinetic Energy:
K.Efinal-K.Einitial
AK. E = [}mcombined (Vground )] – [}mcombined (Vbridge )]
A K. E = [ x 134. 9 × 23. 1877² ] – [ × 134. 9 × 0² ]
A K. E = 36265. 8 J](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b1c4990-3533-4b09-a82c-0110ba869376%2F8322248f-53d7-4fc0-b171-8c5653212865%2Fkxvm768_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The formula of mechanical energy :
М.Е 3D К. Е + Р.E
M.E = mv? + mgh
Where,
m = mass of an object
v = velocity of that object
g = acceleration due to gravity
h = height of an object.
Conservation of Mechanical Energy :
The total mechanical energy of a system is conserved
i. e. the energy can neither be created nor be destroyed.
So,
Mechanical energy at the bridge = Mechanical energy at the ground
M.Epridge = M.Eground
mcombined (v² bridge) + mcombined × g × (hbridge) = ; mcombinedv ground
+ mcombined X g Xx (hground)
x 134.9 x 0 + 134.9 × 9. 8 x 27.432 = x 134.9 x v² + 134.9 × 9. 8
134.9 x 9. 8 x 27. 432 = × 134. 9 × v²
v = 2x134.9x9.8x27432
134.9
V ground = V
2x134.9x9 8x27.432 mls
134.9
Vgroand = 23. 1877 m/s
(b) The total work done by gravity on the two players :
Total work done by gravity on the two players = Loss in Potential energy = Gain in Kinetic energy
AP.E = [mcombined × g × (hbridge)] – [mcombined X g × (hground)]
A P.E = [134.9 × 9. 8 × 27. 432 ] – [134.9 × 9. 8 × 0]
AP.E = 36265, 8 J
AK.E = [mcombined (Vground)] – [¿mcombined (Vbridge )]
A K. E = [ x 134. 9 × 23. 1877² ] – [ x 134. 9 x 0² ]
AK.E = 36265, 8 J
So Work Done by gravity = + 36265, 8 J
(c) The value of final Kinetic Energy:
K.Efinal-K.Einitial
AK. E = [}mcombined (Vground )] – [}mcombined (Vbridge )]
A K. E = [ x 134. 9 × 23. 1877² ] – [ × 134. 9 × 0² ]
A K. E = 36265. 8 J
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