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SOLUTION OF ASSIGNMENT NO. ( 6 )

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

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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 . 0 L L ) 887 ( 2 . 0 30 cos 800 20 cos 1500 ) 6 ( ) 100 ( 2 1 W v m 2 1 v m 2 1 Newton 887 N 0 30 sin 800 1000 20 sin 1500 N : 0 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 0 v A = s / m 6 v B = x y SOLUTION

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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: 0 F y = ( ) 0 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 . 0 s 1 ln 3 150 + + + = 5 13 ln 150 ) 8 ( 20 25 . 0
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SOLUTION OF ASSIGNMENT NO. ( 6 ) - 1 SOLUTION OF ASSIGNMENT...

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