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2 years ago

11

A block of mass of 10kg is being pushed on a tabletop downwards by a force of 200N. There is no acceleration in the problem. Cal

culate the normal force being exerted on the block.

1 answer:

lord [1]2 years ago

7 0

F = M*A

200N*10KG is 2000 newtons

so 2000 newton's is the normal force that is being exerted on the block

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Answer:

Part a)

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Part b)

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Part c)

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Part d)

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Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

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now we have

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$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

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$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

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$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

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