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

To find the block's velocity right before impact, we apply the conservation of energy principle. The potential energy the block has due to its height is fully converted into kinetic energy, allowing us to calculate its speed as approximately 8.8 m/s.

Step-by-step explanation:

The problem involves converting gravitational potential energy into kinetic energy as the 1.50-kg block falls. We can use the conservation of energy principle to solve for the speed of the block just before it hits the floor. The potential energy (PE) at the initial height (h) will be equal to the kinetic energy (KE) just before the block strikes the floor:

PEinitial = KEfinal

mgh = ½ mv2

Solving for speed (v), we have:

v = √(2gh)

Plugging in the values (g = 9.8 m/s2, h = 1.99 m), we get:

v = √(2 * 9.8 m/s2 * 1.99 m)

v = √(39.204 m2/s2 * 1.99 m)

v = √(78.016 m2/s2)

v ≈ 8.8 m/s

Therefore, the block's speed just before it strikes the floor is approximately 8.8 m/s.

Answer 2

6.25 m/s

Step-by-step 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* 9.8* 1.99`

`v^2=2* 9.8* 1.99`

`v=√(2* 9.8* 1.99)`

`v=6.25 m/s`