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2 years ago

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A 161 lb block travels down a 30° inclined plane with initial velocity of 10 ft/s. If the coefficient of friction is 0.2, the to

tal work done, in ft-lbs by all forces when the block moves through a distance of 5 feet most nearly is?

1 answer:

lbvjy [14]2 years ago

5 0

Answer:

8418 ft lbf

Explanation:

Let g = 32 ft/s2. The gravitational force that is parallel to the incline is

$mgsin30^o = 161* 32* 0.5 = 2576 lbf$

The normal force that acting on the block is the gravity force component that is perpendicular to the incline:

$N = mgcos30^o = 32*161*\sqrt{3}/2 \approx = 4461.8 lbf$

The friction force is the product of normal force and coefficient

$F_f = \mu N = 0.2*4461.8 = 892.4 lbf$

So the net force is force of gravity subtracted by friction force

F = 2576 - 892.4 = 1683.6 lbf

So the work by this force over 5 feet distance is

W = Fs = 1683.6*5 = 8418 ft lbf

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