Question

A1 does kilogram block slides down a frictionless inclined plane 2.5 M High what is the velocity of the bottom of the plane assume gravitational acceleration is 9.8 Ms is it 2.5 mm is 3.9 m s 7.0 m s r 9.8 m s?

Answer 1

The velocity is 7.0m/s

Step-by-step explanation:

Given

`mass (m) = 1kg`

`height (h) = 2.5m`

`g = 9.8m/s^2`

Required

The bottom of the plane velocity

To do this, we apply the work-energy theorem which states that the energy at the highest point and at the lowest point are equal.

At the highest point

`v = 0`

`E = (1)/(2)mv^2 + mgh`

`E = (1)/(2)*1*0^2+ 1 * 9.8 * 2.5`

`E = 0 + 24.5`

`E = 24.5`

At the lowest point

`h = 0`

`E = (1)/(2)mv^2 + mgh`

`E = (1)/(2) * 1 * v^2 + mg(0)`

`E = (1)/(2) * 1 * v^2 + 0`

`E = (v^2)/(2)`

Equate both values of energy

`24.5 = (v^2)/(2)`

`v^2 = 2 * 24.5`

`v^2 = 49`

Take square roots of both sides

`v = 7`

The velocity is 7.0m/s