Modelling motion — Question 270

ANSWER

(a) Weight 40 N vertically downward; reaction normal away from the plane; tension 18 N at $20^\circ$ above the uphill direction. (b) $R=40\cos30^\circ-18\sin20^\circ\approx28.48$ N; acceleration $0.771$ m s$^{-2}$ downhill.

Worked solution

(a) The block experiences gravity, the plane reaction and rope tension. A smooth plane has no friction force. See the solution diagram. Resolve components relative to the plane, not relative to the rope.

(b) Normal acceleration is zero, so $R+18\sin20^\circ=40\cos30^\circ$. Uphill force is $18\cos20^\circ-40\sin30^\circ\approx-3.0855$ N. Dividing by 4 gives $a\approx-0.7714$ m s$^{-2}$ uphill, i.e. $0.771$ m s$^{-2}$ downhill (three decimals). A positive reaction confirms continued contact.

Equivalent justified methods accepted; respect any requested proof method.

Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.

Part a: Draw and label weight vertically downward.

[1 mark]

Part a: Draw and label the normal reaction perpendicular to the plane.

[1 mark]

Part a: Draw tension in the stated direction from the centre node.

[1 mark]

Part b: Resolve normal forces and obtain R.

[1 mark]

Part b: Resolve the net along-plane force.

[1 mark]

Part b: Divide by mass and state downhill acceleration.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.