Asked by gale gale
A body of mass 11kg falls freely from rest from a height of 15m above the ground. 3m above the ground it breaks through a thin roof before continuing its fall.
(a) given that the work done in breaking through the roof is 130 J. Calculate the velocity of the body after it emerges from the roof.
(b) calculate the velocity of the body as it hits the ground.
(c) An identical body is dropped from the same height through the hole in the roof made by the first body . Ccalculate the velocity of the b0dy sas it hits the ground
(d) which mechanical principle did you apply when answering the above.
(a) given that the work done in breaking through the roof is 130 J. Calculate the velocity of the body after it emerges from the roof.
(b) calculate the velocity of the body as it hits the ground.
(c) An identical body is dropped from the same height through the hole in the roof made by the first body . Ccalculate the velocity of the b0dy sas it hits the ground
(d) which mechanical principle did you apply when answering the above.
Answers
Answered by
Henry
a. KE = 0.5*M*V^2 = 130 J.
0.5*11*V^2 = 130
V^2 = 23.64
V = 4.86 m/s.
b. V^2 = Vo^2 + 2g*h = 4.86^2 + 19.6*3 =
82.4
V = 9.1 m/s
c. V^2 = Vo^2 + 2g*h = 0 + 19.6*15 = 294
V = 17.15 m/s.
0.5*11*V^2 = 130
V^2 = 23.64
V = 4.86 m/s.
b. V^2 = Vo^2 + 2g*h = 4.86^2 + 19.6*3 =
82.4
V = 9.1 m/s
c. V^2 = Vo^2 + 2g*h = 0 + 19.6*15 = 294
V = 17.15 m/s.
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