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3 years ago

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1. Suppose a 10 N force is applied to the side of a 4.0 kg block that is sitting on a table what would be the minimum value of t

he coefficient of static friction in order for the block to remain motionless?

1 answer:

vlada-n [284]3 years ago

3 0

Answer:

$\mu= 0.25$

Explanation:

Given data

Force= 10N

mass= 4kg

r= 4*9.81

r= 39.24N

The expression for the force acting is expressed as

$F=\mu*r$

substitute

$\mu=F/r$

$\mu= 10/39.24$

$\mu= 0.25$

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4 years ago

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Answer:

Part a)

$t = 1.65 s$

Part b)

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Part c)

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Explanation:

Part a)

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So the acceleration due to gravity on this new planet is given as

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now by kinematics we have

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Part b)

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now the distance moved by it

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Part c)

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now magnitude of velocity is given as

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$v = 27.3 m/s$

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Answer:

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