A 17.9 kg mass, slides down a rough inclined plane at an angle of 33.7 to the horizontal, experiencing a frictional force of 5.6 N.Determine the net force acting on the mass.

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First, identify the forces acting on the mass along the incline. The weight of the mass is W = m g. With m = 17.9 kg and g ≈ 9.8 m/s^2, W ≈ 17.9 × 9.8 = 175.42 N.
Next, resolve this weight into a component parallel to the incline: F_parallel = W sin(θ), where θ = 33.7°. sin(33.7°) ≈ 0.556 (approximate value). Thus F_parallel ≈ 175.42 × 0.556 ≈ 97.6 N (this is the component pulling the mass down the plane).
The frictional force opposes the motion and has magnitude f = 5.6 N, acting up the incline.
Apply the net force along the incline: F_net = F_parallel − f ≈ 97.6 N − 5.6 N ≈ 92.0 N.
Therefore, the net force acting on the mass along the incline is about 92 N in the downward direction along the plane.
If a given answer is 86.72 N, this value would correspond to a different sine value, a different g, or an arithmetic miscalculation. Using standard g ≈ 9.8 m/s^2 and sin(33.7°) ≈ 0.556 yields ~92 N, not ~86.7 N. The discrepancy highlights the sensitivity to the sine value and rounding during intermediate steps.
Summary: compute the downhill weight component along the plane, subtract the uphill friction, and interpret the sign/direction as down the plane; with the provided numbers the result is approximately 92 N, not 86.72 N.

MUF0121 Physics Unit 1 - Semester 2, 2025

A 17.9 kg mass, slides down a rough inclined plane at an angle of 33.7 to the horizontal, experiencing a frictional force of 5.6 N.Determine the net force acting on the mass.

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