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Relation between Work Done and PE

If m1 = 20k...

Question

If $m_{1} = 20 k g & m_{2} = 10 k g$ find the normal force acting on the floor as shown in fig.?

A

300 N

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B

100 N

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C

200 N

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D

None of these

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Solution

The correct option is C 300 N
Given:

$m_{1} = 20 k g$
$m_{2} = 10 k g$

$g = 10 m / s^{2}$
Normal force is equal to the weight of system, it can be given as! $N_{F} = (m_{1} + m_{2}) g$
$N = (20 + 10) 10$
$N_{F} = 300 N$
[A] $300 N$

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