a ball of mass 200 g falls from a height of 5m. what will be its kinetic energy when it first reaches the ground
sumanta22:
here m=200g, we can convert it into si system it means (200×1000)kg =200000, h=5m andg =9.8 N then we can do according to the math m×g×h=200×1000×5×9.8=9800000jule.
its wrong
1kg=1000g
so 1g=1/1000
kg
200g=200/1000kg
=>200g=0.2g or 2/10g
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Use energy conservation law.
Total Energy at height 5m = total energy just before hitting ground
Potential energy at height 5m = mgh = 0.2kg × 10 × 5 = 10 J
KE = 0 as it was at rest.
Just before hitting the ground,
PE = mgh = mg × 0 = 0
Let Kinetic energy = KE'
Now Total Energy at height 5m = total energy just before hitting ground
=> 10 + 0 = 0 + KE'
=> KE' = 10J
Just before hitting the ground, KE = 10J
Total Energy at height 5m = total energy just before hitting ground
Potential energy at height 5m = mgh = 0.2kg × 10 × 5 = 10 J
KE = 0 as it was at rest.
Just before hitting the ground,
PE = mgh = mg × 0 = 0
Let Kinetic energy = KE'
Now Total Energy at height 5m = total energy just before hitting ground
=> 10 + 0 = 0 + KE'
=> KE' = 10J
Just before hitting the ground, KE = 10J
Awesome
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