A block of weight `15N` slides on a horizontal table the co-efficient of siliding fricition is `0.4`. The area of the block in contact with the table is `0.5m^(2)`. The shearing stress will beA. `120 Nm^(-2)`B. `140 Nm^(-2)`C. `160 Nm^(-2)`D. `180 Nm^(-2)`

A block of weight `15N` slides on a horizontal table the co-efficient

1.	A block of weight `15N` slides on a horizontal table the co-efficient of siliding fricition is `0.4`. The area of the block in contact with the table is `0.5m^(2)`. The shearing stress will beA. `120 Nm^(-2)`B. `140 Nm^(-2)`C. `160 Nm^(-2)`D. `180 Nm^(-2)`
Answer» Correct Answer - A Stress `= (mu mg)/(A) = F = f = mu mg`

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A block of weight `15N` slides on a horizontal table the co-efficient of siliding fricition is `0.4`. The area of the block in contact with the table is `0.5m^(2)`. The shearing stress will beA. `120 Nm^(-2)`B. `140 Nm^(-2)`C. `160 Nm^(-2)`D. `180 Nm^(-2)`

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