A block has been placed on an inclined plane. the slope angle θ of the plane is such that the block slides down the plane at a constant speed. the coefficient of kinetic friction is equal to
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Assuming the block's mass is m, the normal reaction of the inclined plane is mgcos© where © is the angle( sorry I don't have theta) of inclination. Let the coefficient of kinetic friction be ¢. Then as the block is moving with constant speed the friction force, ¢mgcos© and mgsin© cancel out. Therefore ¢mgcos©=mgsin© which implies that ¢=tan©
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fk = meiun(k) mg cos/thitha and acceleration = 0 (since speed is constant)
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