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

Answer:

The velocity is 7.0m/s

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

$v = 0$

$E = \frac{1}{2}mv^2 + mgh$

$E = \frac{1}{2}*1*0^2+ 1 * 9.8 * 2.5$

$E = 0 + 24.5$

$E = 24.5$

At the lowest point

$h = 0$

$E = \frac{1}{2}mv^2 + mgh$

$E = \frac{1}{2} * 1 * v^2 + mg(0)$

$E = \frac{1}{2} * 1 * v^2 + 0$

$E = \frac{v^2}{2}$

Equate both values of energy

$24.5 = \frac{v^2}{2}$

$v^2 = 2 * 24.5$

$v^2 = 49$

Take square roots of both sides

$v = 7$

The velocity is 7.0m/s