A man drops a 10 kg rock from the top of a 5m ladder .What is it's speed just before it hits the ground ? What is the its kinetic energy when it's reaches the ground?
Potential energy at the top=mgh =10×10×5 (g=10m/s) =500J kinetic energy when it reaches the ground=1/2×mv^2 potential energy=kinetic energy(law of conservation of energy) 500J=500J °•°kinetic energy=500J the speed of the rock before reaching the ground=1/2×10×v^2=500 =5×v^2=500 v^2=500/5 v^2=100 v=10m/s
Answers & Comments
Verified answer
Potential energy at the top=m*g*h=10X10X5 (considering g=10m/s)
=500J
kinetic energy when it reaches the ground=1/2mv²
=1/2*10*v²
as we know,
potential energy=kinetic energy (law of conservation of energy)
500J=500J
∴kinetic energy = 500J
the speed of rock before reaching the ground is
1/2*10*v²=500
5*v²=500
v=500/5
v=10m/s
Verified answer
Potential energy at the top=mgh=10×10×5
(g=10m/s)
=500J
kinetic energy when it reaches the ground=1/2×mv^2
potential energy=kinetic energy(law of conservation of energy)
500J=500J
°•°kinetic energy=500J
the speed of the rock before reaching the ground=1/2×10×v^2=500
=5×v^2=500
v^2=500/5
v^2=100
v=10m/s