Determine the coefficient of dynamic friction between the stone and the ice during the last 14.0 s of the stone’s motion.

The diagram shows the stone during its motion after release.

Label the diagram to show the forces acting on the stone. Your answer should include the name, the direction and point of application of each force.

ALTERNATIVE 1

«deceleration» \( = \frac{{3.41}}{{14.0}}\) «\( = 0.243\,{\text{m}}\,{{\text{s}}^{ - 2}}\)»

F = 0.243 × m

\(\mu  = \frac{{0.243 \times m}}{{m \times 9.81}} = 0.025\)

ALTERNATIVE 2

distance travelled after release = 23.85 «m»
KE lost = 5.81m «J»

\({\mu _{\text{d}}} = \frac{{{\text{KE lost}}}}{{mg \times {\text{distance}}}} = \frac{{5.81m}}{{23.85mg}} = 0.025\)

Award [3] for a bald correct answer.

Ignore sign in acceleration.

Allow ECF from (a) (note that \(\mu  = 0.0073\) x candidate answer to (a) ).

Ignore any units in answer.

Condone omission of m in solution.

Allow g = 10 N kg^–1 (gives 0.024).

normal force, upwards, ignore point of application

Force must be labeled for its mark to be awarded. Blob at poa not required.
Allow OWTTE for normal force.  Allow N, R, reaction.
The vertical forces must lie within the middle third of the stone

weight/weight force/force of gravity, downwards, ignore point of application

Allow mg, W but not “gravity”.

Penalise gross deviations from vertical/horizontal once only

friction/resistive force, to left, at bottom of stone, point of application must be on the interface between ice and stone

Allow F, μR. Only allow arrows/lines that lie on the interface. Take the tail of the arrow as the definitive point of application and expect line to be drawn horizontal.

Award [2 max] if any force arrow does not touch the stone

Do not award MP3 if a “driving force” is shown acting to the right. This need not be labelled to disqualify the mark. Treat arrows labelled “air resistance” as neutral.

N.B: Diagram in MS is drawn with the vertical forces not direction of travel collinear for clarity