Answer:
Explanation:
Height of the block before the release "h" = 1.99 kg
mass of the block "m" = 1.5 kg
potential energy of the block = m g h
= 1.5 x 9.8 x 1.99
= 29.253 J
If v be its velocity before striking the floor
kinetic energy of block = 1/2 m v²
Applying conservation of mechanical energy
kinetic energy = potential energy
1/2 m v² = 29.253
.5 x 1.5 x v² = 29.253
v² = 39
v = 6.24 m /s
Answer:
6.25 m/s
Explanation:
We are given that
Mass of block,m=1.5 kg
Distance,s=1.99 m
Initial velocity,u=0
We have to find the speed just before it strikes the floor.
We know that
[tex]v^2-u^2=2gs[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Substitute the values
[tex]v^2-0=2\times 9.8\times 1.99[/tex]
[tex]v^2=2\times 9.8\times 1.99[/tex]
[tex]v=\sqrt{2\times 9.8\times 1.99}[/tex]
[tex]v=6.25 m/s[/tex]
