A block of weight 200N is pulled along a rough horizontal surface at constant speed by a force 100N acting at an angle $30^{0}$ above the horizontal. The coefficient of kinetic friction between the block and the surface is:

0.43

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0.58

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0.75

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0.83

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Solution

The correct option is D 0.58
For vertical equilibrium, $N = 200 - 100 s i n 30^{0} = 150 N$
For horizontal equilibrium, $f = 100 c o s 30^{0}$
$f = μ N$
$μ = \frac{100 c o s 30^{0}}{150}$
$μ = 0.58$

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A block of weight 200N is pulled along a rough horizontal surface at constant speed by a force 100N acting at an angle 300 above the horizontal. The coefficient of kinetic friction between the block and the surface is:

The correct option is D 0.58For vertical equilibrium, N=200−100sin300=150NFor horizontal equilibrium, f=100cos300f=μNμ=100cos300150μ=0.58

A block of weight 200N is pulled along a rough horizontal surface at constant speed by a force 100N acting at an angle $30^{0}$ above the horizontal. The coefficient of kinetic friction between the block and the surface is:

The correct option is D 0.58
For vertical equilibrium, $N = 200 - 100 s i n 30^{0} = 150 N$
For horizontal equilibrium, $f = 100 c o s 30^{0}$
$f = μ N$
$μ = \frac{100 c o s 30^{0}}{150}$
$μ = 0.58$