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?

icon
Related questions
Question
100%
This is phyiscs question i attached the soloution also. I don't need this soloution that is from chegg and this is wrong. Plz provide me a clear handwritten and perfect soloution in one hour and take a thumb up plz
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?
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 >
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 >
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer