SOLUTION OF ASSIGNMENT NO. ( 6 )

SOLUTION OF ASSIGNMENT NO. ( 6 ) - 1 1 SOLUTION OF...

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Unformatted text preview: 1 1 SOLUTION OF ASSIGNMENT NO. ( 6 ) PRINCIPLE OF WORK AND ENERGY 1. The 100–kg crate is subjected to two forces F 1 and F 2 as shown. If it is originally at rest , determine the distance it slides in order to attain a speed of v = 6 m / s. The coefficient of kinetic friction between the crate and the surface is μ = 0.2 . [ ] m 934 . L L ) 887 ( 2 . 30 cos 800 20 cos 1500 ) 6 ( ) 100 ( 2 1 W v m 2 1 v m 2 1 Newton 887 N 30 sin 800 1000 20 sin 1500 N : F o o 2 B A 2 A 2 B o o y = − + = ⇒ = − = ∴ = − − + = → ∑ ° 20 ° 30 1500N 800N No.(1) 1000 N N L 1500 cos ° 20 1500 sin ° 20 800 sin ° 30 800 cos ° 30 N μ A B v A = s / m 6 v B = x y SOLUTION 2 2 2. The 2–kg block is subjected to a force having a constant direction and a magnitude which is given by F = 300 / ( 1 + s ) N , where s is in meters.When s = 4 m , the block is moving to the right with a speed of 8 m / s . Determine its speed when s = 12 m . The coefficient of kinetic friction between the block and the ground is μ = 0.25 . SOLUTION: F y = ∑ ( ) 20 s 1 30 sin 300 N = − + ° − ( ) s 1 150 20 N + + = (1) ∑ → = − 2 1 2 o 2 W mv 2 1 mv 2 1 ( ) ∫ ∫ ∑ − + ° = → 12 4 12 4 2 1 s d N s d s 1 30 cos 300 W μ Substituting N from equation (1): ( ) ( ) s d s 1 150 20 s d s 1 30 cos 300 W 12 4 12 4 2 1 ∫ ∫ ∑ + + − + ° = → μ = ( ) [ ] ( ) [ ] 12 4 12 4 s 1 ln 150 s 20 25 . s 1 ln 3 150 + + − + − −...
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This note was uploaded on 12/21/2010 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.

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SOLUTION OF ASSIGNMENT NO. ( 6 ) - 1 1 SOLUTION OF...

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