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The block is always at rest, the maximum...

The block is always at rest, the maximum force which can be applied for this is 24x. Find the value of x. Take g = 10 `m//s^(2)`{coefficient of friction = 0.3}

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The correct Answer is:
`0-25`
`f_(max)=mu m g`
`=3xx2xx10 =6N =24xx1/4 therefore x=.25`
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