IB Physics HL A.2.7 Friction and Resistive Forces

Practise analysing HL friction systems with changing surfaces, coupled blocks, circular motion, slopes and drag.

Exam points

Analyse static and dynamic friction in coupled blocks and changing-force systems.
Apply friction, drag and circular-motion models to calculate limiting conditions.

[Maximum number: 5]

A girl on a sledge is moving down a snow slope at a uniform speed.

Figure for Question A.2.7—Frictional force question 1 — IB Physics HL

(a)

The sledge, without the girl on it, now travels up a snow slope that makes an angle of $6.5^{\circ}$ to the horizontal. At the start of the slope, the speed of the sledge is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The coefficient of dynamic friction of the sledge on the snow is 0.11 .

[ 3 ]

(i)

Show that the acceleration of the sledge is about $-2 \mathrm{~m} \mathrm{~s}^{-2}$ .

[ 3 ]

«friction force down slope» $=\mu m g \cos (6.5)=$ «5.9N» «component of weight down slope» $=m g \sin (6.5)$ « $=6.1 \mathrm{~N}$ »
$\text { «so } a=\frac{F}{m} \text { » acceleration }=\frac{12}{5.5}=2.2 \text { « } \mathrm{ms}^{-2} \text { » }$

Marking guidance:

Ignore negative signs Allow use of $g=10 \mathrm{~ms}^{-2}$

(b)

The coefficient of static friction between the sledge and the snow is 0.14 . Outline, with a calculation, the subsequent motion of the sledge.

[ 2 ]

calculates a maximum value for the frictional force = « $\mu R=$ » 7.5 « N » sledge will not move as the maximum static friction force is greater than the component of weight down the slope

Marking guidance:

Allow correct conclusion from incorrect MP1 Allow 7.5 > 6.1 so will not move

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A.2.7—Frictional force question 1

Question (a)

Question (i)

Question (b)