a body of 2 kg fall from rest what will be its kinetic energy during the fall at the end of2 second
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Answer:
The answer is 400J
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Answer:
400 J
Explanation:
KE = 1/2mv*v
h = height
u = initial velocity
g = acceleration due to gravity
t = time taken
By 2nd law of motion,
h = ut + 1/2gt*t
h = (0*2) + 1/2*10*2*2
h = 20 m
By 3rd equation of motion,
v² = u² + 2gh
⇒ v² = 0² + 2*10*20
⇒ v² = 400
⇒ v = 20
KE = 1/2 × 2× 20²
= 400 J
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