A force of 98 newton is required to pull a body of mass 1×102 kg over the surface of the ice then the co-efficient of friction is (g=10ms−2)

f=98N,m=1×102kg μms=? fms=μsN fms=μsmg 98=μs×1×102×9.8 μs=98980 μs=0.1

A force of 98 newton is required to pull a body of mass 1×102 kg over the surface of the ice then the co-efficient of friction is (g=10ms−2)

f=98N,m=1×102kg μms=? fms=μsN fms=μsmg 98=μs×1×102×9.8 μs=98980 μs=0.1

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A force of 98 newton is required to pull a body of mass $1 \times 10^{2} kg$ over the surface of the ice then the co-efficient of friction is $(g = 10 {ms}^{- 2})$

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Detailed Solution

f = 98 N,m = 1 \times 10^{2} kg

$μ_{ms} = ?$

$f_{ms} = μ_{s} N$

$f_{ms} = μ_{s} mg$

$98 = μ_{s} \times 1 \times 10^{2} \times 9.8$

$μ_{s} = \frac{98}{980}$

$μ_{s} = 0.1$

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