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प्रश्न
A body of mass 10 Kg is kept at a height of 5m. It is allowed to fall and reach the ground.
(i) What is the total mechanical energy possessed by the body at the height of 2m assuming it is a frictionless medium ?
(ii) What is the kinetic energy possessed by the body just before hitting the ground ? Take g = 10 m/s2.
उत्तर
m = 10 kg
h = 5 m
g = 10 m/s²
Mechanical Energy = Potential Energy + Kinetic Energy
Potential Energy at h of 5 m = mgh = (10) ×(10) × (5) = 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 = 10`m/(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 = 10`m/(s²)`
=> 2×10×3 = V²
=> V² = 60
Kinetic Energy = `(1/2) `10 × 60 = 300 J
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