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stealth61 [152]
2 years ago
9

a 77 N net force is applied to a box which slides horizontally across a floor for 6.7 m. What amount of work is done on the box

by the net force ?
Physics
1 answer:
ArbitrLikvidat [17]2 years ago
7 0
W=fd
W=77*6.7
W=515.9 J
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A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio
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Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

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