Question

A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. if the block is released from rest at a distance of 1.99 m above the floor, what is its speed just before it strikes the floor?

Answer 1

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 2

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

$v^2-u^2=2gs$

Where $g=9.8 m/s^2$

Substitute the values

$v^2-0=2\times 9.8\times 1.99$

$v^2=2\times 9.8\times 1.99$

$v=\sqrt{2\times 9.8\times 1.99}$

$v=6.25 m/s$