Question #15643

a boy sitting on a swing at a maximum height of 5m above the ground.when the swing passes through the mean position which is 2m above the ground its velocity is?

Expert's answer

The law of conservation of energy states that the total amount of

energy in

an isolated system remains constant over time.

At a maximum height:

K0 =

0

P0 = m * g * H

At a lowest point:

K1 = m * V^2 / 2

P1 = m * g *

h

K0 + P0 = K1 + P1, so:

m * g * H = m * V^2 / 2 + m * g * h

g * H

= V^2 / 2 + g * h

V^2 / 2 = g * H - g * h

V = sqrt(2 * g * (H - h))

V =

sqrt(2 * 9.81 * (5 - 2)) = 7.67 m/s

energy in

an isolated system remains constant over time.

At a maximum height:

K0 =

0

P0 = m * g * H

At a lowest point:

K1 = m * V^2 / 2

P1 = m * g *

h

K0 + P0 = K1 + P1, so:

m * g * H = m * V^2 / 2 + m * g * h

g * H

= V^2 / 2 + g * h

V^2 / 2 = g * H - g * h

V = sqrt(2 * g * (H - h))

V =

sqrt(2 * 9.81 * (5 - 2)) = 7.67 m/s

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