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A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acting on the block? Round your answer to the nearest whole number.

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akindelemf

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ =  30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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