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3 years ago

A box of mass m is held at rest on a frictionless surface with force F up the ramp. The ramp has an angle

Physics

2 answers:

frutty [35]3 years ago

8 0

Answer:

Explanation:

Andrew [12]3 years ago

7 0

Answer:

mg cos θ

Explanation:

correct answer on khan academy

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2 years ago

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

Dmitry [639]

Answer:

Part a)

$F = 135.7 N$

Part b)

$F = 62.5 N$

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

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3 years ago

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Answer:

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Explanation:

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Answer:

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