A man drops a 10kg rock from the top of a 5m ladder.What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground?
( g = 9.8 m/s )
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Answered by
9
By conservation of energy
Potential energy at the top = Kinetic energy at the ground
mgh= 1/2 m*v²
v²= 2gh
v²= 2*9.8*5
v²= 98
v= √98 m/s
K.E= 1/2 *m*v²
K.E.= 1/2* 10* 98
K.E.= 490 J
Potential energy at the top = Kinetic energy at the ground
mgh= 1/2 m*v²
v²= 2gh
v²= 2*9.8*5
v²= 98
v= √98 m/s
K.E= 1/2 *m*v²
K.E.= 1/2* 10* 98
K.E.= 490 J
Dhruv00:
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But answer is speed = 10m/s and KE = 500J
thats same but if u take g= 10 m/s² then ull get speed = 10m/s ,but u have given the value of g as 9.8m/s²
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