# Question 5:

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram with four forces showing arrow heads and labelled | B1 | Allow $mg$ or $8g$; allow $T$ or 40 or other reasonable notation; allow $\mu R$; direction of friction must be to the left; any components must be shown in a different style |

## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8g + 40\sin 30° (= R)$ | M1 | Expression for normal reaction, with $mg$ or $8g$ and $40\sin 30°$ or $40\cos 30°$; allow incorrect signs |
| Correct expression with correct signs | A1 | |
| $R = 98.4 \text{ N}$ | A1 | Correct value from correct working; use of $g = 9.81$ gives 98.5 N; do not penalise if already done so earlier in script; otherwise penalise by 1 mark |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F = 40\cos 30° = 34.6 \text{ N}$ | M1 | Use of $40\cos 30°$ or $40\sin 30°$; award M0 if any extra terms |
| Correct value for friction | A1 | Don't need to see $F$ |

## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $40\cos 30° \leq \mu \times 98.4$ | M1 | Use of $F \leq \mu R$ (or $F = \mu R$); must use $R = 98.4$ and a positive value for $F$ |
| Correct inequality or equation | A1F | Allow use of $F = \mu R$ throughout |
| $\mu \geq \frac{40\cos 30°}{98.4}$, $\mu \geq 0.352$ | A1F | Correct minimum value; follow through must use $R = 98.4$ and their $F$ from (c); use of $\sin 30°$ in part (c) gives 0.203 |

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