Gravity and motion — Question 247

ANSWER

(a) Weight vertically down; normal force perpendicular away from the plane; tension and friction uphill. (b) $2.90\,\mathrm{m\,s^{-2}}$ downhill.

Worked solution

(a) The four arrows represent real interactions. Friction opposes the downhill sliding. The normal force is perpendicular to the plane, rather than vertical.

(b) $F_{\rm net}=mg\sin30^\circ-T-f=9.80-3.00-1.00=5.80\,\mathrm N$. Hence $a=5.80/2.00=2.90\,\mathrm{m\,s^{-2}}$.

Equivalent physically justified methods, alternative valid sketches and consistent units accepted.

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Draw weight vertically down from the node.

[1 mark]

Part a: Draw the normal force perpendicular to the plane from the node.

[1 mark]

Part a: Draw and distinguish uphill tension and uphill friction from the node.

[1 mark]

Part b: Resolve weight and determine the signed net force.

[1 mark]

Part b: Divide by mass and state downhill.

[1 mark]

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