2) Assume that energy level needed for skull fracture is 50 ft-lbs. a) What is the minimum height that a 2 lb chuck of metal must fall to cause skull fracture? b) What is its velocity at impact?
2) Assume that energy level needed for skull fracture is 50 ft-lbs. a) What is the minimum height that a 2 lb chuck of metal must fall to cause skull fracture? b) What is its velocity at impact?
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![2) Assume that energy level needed for skull fracture is 50 ft-lbs.
a) What is the minimum height that a 2 lb chuck of metal must fall to cause skull fracture?
b) What is its velocity at impact?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffab38f68-54cf-417c-ad7b-664663ca6fc0%2Ffbc13fa4-24c8-44e2-80d4-a1cda975a86b%2Fmu151uw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2) Assume that energy level needed for skull fracture is 50 ft-lbs.
a) What is the minimum height that a 2 lb chuck of metal must fall to cause skull fracture?
b) What is its velocity at impact?
![Anonymous answered this
6,949 answers
(a) Suppose the minimum height of the chuck of metal is 'h' ft.
So, potential energy of the metal chuck, PE = m*g*h
this must be equal to 50 ft-lbs
So, m*g*h = 50
=> 2*32.2*h = 50
=>h = 50 / (2*32.2) = 0.78 ft.
Therefore, the minimum height = 0.78 ft.
(b) Suppose the velocity of impact = v ft/s.
So, Kinetic Energy of the metal chuck, KE = (1/2)*m*v^2
this must be equal to 50 ft-lbs.
So, 0.5*2*v^2 = 50
=> v = V50 = 7.1 ft/s.
Comment >](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffab38f68-54cf-417c-ad7b-664663ca6fc0%2Ffbc13fa4-24c8-44e2-80d4-a1cda975a86b%2Feyokj8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Anonymous answered this
6,949 answers
(a) Suppose the minimum height of the chuck of metal is 'h' ft.
So, potential energy of the metal chuck, PE = m*g*h
this must be equal to 50 ft-lbs
So, m*g*h = 50
=> 2*32.2*h = 50
=>h = 50 / (2*32.2) = 0.78 ft.
Therefore, the minimum height = 0.78 ft.
(b) Suppose the velocity of impact = v ft/s.
So, Kinetic Energy of the metal chuck, KE = (1/2)*m*v^2
this must be equal to 50 ft-lbs.
So, 0.5*2*v^2 = 50
=> v = V50 = 7.1 ft/s.
Comment >
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