Question

A light string is wrapped around the edge of the smaller disk, and a 1.50 kgkg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.40 mm above the floor, what is its speed just before it strikes the floor

Answer 1

Answer:

The speed of the block before it strikes the floor is 0.217 m/s.

Explanation:

Given;

mass of the block, m = 1.5 kg

height above the ground through the block was released, h = 2.4 mm = 2.4 x 10⁻³ m

The speed of the block before it strikes the floor will be maximum.

Let the speed of the block before it strikes the floor = v

Apply the principle of conservation of mechanical energy to determine the speed of the block.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 2.4 x 10⁻³)

v = 0.217 m/s

Therefore, the speed of the block before it strikes the floor is 0.217 m/s.