a block of mass 5 kg is placed on a rough inclined surface of inclination 30 degree with horizontal if the block is at rest on inclined surface then the force exerted by inclined surface on block is1)25 N2)25root3N3)50N4)100N
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Downward force on an object when force is on rough inclined surface is given as:
F = mg sin Ф
F = 5 * 10 sin 60
F = 50 * √3 / 2
F = 25 √3 N
F = mg sin Ф
F = 5 * 10 sin 60
F = 50 * √3 / 2
F = 25 √3 N
arkatrishadas1pec2j6:
answer is 50
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25 N is the answer bcoz
mgsin(theta) force box will experience
so F = 5 x 10 x sin30
= 50x1/2 = 25 N.
mgsin(theta) force box will experience
so F = 5 x 10 x sin30
= 50x1/2 = 25 N.
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