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.

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