# A man drops a 10kg rock from the top of a 5m ladder. What is its kinetic energy when it reaches the ground ?.what is its speed just before it hits the ground?

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by BCM

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by BCM

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mechanical energy=kinetic +potential

(mass=10kg,g=10m/s^2,h=5m)

potential energy at the top(maximum)=mgh

=10X10X5

=500J

kinetic energy when it reaches the ground(maximum)=1/2mv^2

=1/2*10*v^2

potential energy=kinetic energy (law of conservation of energy)

500J=500J

therefore,kinetic energy=500J

THE SPEED JUST BEFORE IT HITS THE GROUND IS-

1/2*10*V^2=500

V^2=500/5

V=10

THEREFORE V=10

Height = 5 m

so, by equation of position velocity relation

v² - u² = 2as

u = 0 m/s

a = g =10 m/s²

s = h = 5m

so, velocity just before hitting the ground = 10 m/s

and the kinetic energy = 1/2 mv² = 500 J