A body of mass 10 kilogram is kept at a height of 5m it is allowed to fall and reach the ground what is the total mechanical energy possessed by the body at a height of 2m assuming it is a frictionless most medium what is the kinetic energy possessed by the body just before hitting the ground
Answers
Mass = 10 kg
Height = 5 m
g = 10 m/s²
Mechanical Energy = Potential Energy + Kinetic Energy
Potential Energy at h of 5 m = mgh = 10 * 10 * 5 = 500
Kinetic energy at h of 5m = 0 as body is at rest (KE = (1/2)mV² and V=0) so KE = 0)
Total Mechanical Energy = 500 + 0 = 500 J
i)
Energy neither can be created or destroyed but can be converter in to another form. as its a Frictionless surface so no energy loss in any other form
so
Total Mechanical Energy at h of 2m = 500 J
ii)
Just Before hitting the ground h = 0 m
so potential energy = 0 J
Total Mechanical Energy = 500 J
500 = 0 + Kinetic Energy
=> Kinetic Energy = 500 J
or we can find it by using
KE = (1/2)mV²
V² - U² = 2aS U = 0 S = 5 , a = g = 10m/s²
V² = 2*10*5= 100
KE = (1/2)*10 * 100 = 500 J
in part i) if we need to found Kinetic energy & Potential energy separately
Potential energy at 2 m = 10 * 10 * 2 = 200 J
Kinetic Energy = Total energy - potential energy = 500 - 200 = 300 J
or we can calculate by
Kinetic Energy = (1/2) m V²
2aS = V² - U²
U = 0 , S = 5 -2 = 3 m a = g = 10m/s²
=> 2*10*3 = V²
=> V² = 60
Kinetic Energy = (1/2) 10 * 60 = 300 J
Answer:
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