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3 years ago

A rope pulls a 82.5 kg skier at a constant speed up a 18.7Â° slope with Î¼k = 0.150. How much force does the rope exert?

Physics

1 answer:

Artist 52 [7]3 years ago

6 0

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f = μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

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2. A jack exerts a vertical force of 4.5 X 103

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Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

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$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

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$\\ \\$

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$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

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$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

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$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

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