A 500kg block is @ an top of 30 degree incline plane .when it is released it slides 140cm down along the plane. what will the speed of the block when it reaches the bottom.

Here,

Mass, m = 500 kg

Inclination of the inclined plane, θ = 30o

Distance traveled along the plane, s = 140 cm = 1.4 m

From the diagram,

AB = s = 1.4 m

AC = AB sinθ = (1.4)(sin 30) = (1.4)(1/2) = 0.7 m

Considering C to be the reference level, the potential energy at A is = mg(AC)

This energy is converted into kinetic energy at B = ½ mv2

[since, C is the reference level, the potential energy at C is zero]

So, by energy conservation,

mg(AC) = ½ mv2

=> (9.8)(0.7) = ½ v2

=> v = 3.7 m/s

This is the velocity at the bottom of the plane.

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use the principal of conversion of potential energy due to gravity into kinetic energy of sliding.

mgh = 1/2 mv2

v2 = 2gh

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but h is vertical height from bottom.

h/140 = sin(30)

h = 140 x sin(30) = 140 x 0.5 = 70 cm = 0.7 m

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v2 = 2gh

v2 = 2x 9.8 x 0.7

v = 3.7 m/s

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