A parcel of mass 5 kg lies on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{1}{4}$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Using the provided value for $\tan \alpha = \frac{1}{4}$, we can find:

• Calculate $\sin \alpha$ and $\cos \alpha$. Since $\tan \alpha = \frac{1}{4}$: $\sin \alpha = \frac{1}{\sqrt{17}}; \quad \cos \alpha = \frac{4}{\sqrt{17}}$

3. Substitute $g = 9.81$ m/s$^2$ and compute:
   • For $R$: $R = 5(9.81) \cdot \frac{4}{\sqrt{17}} + 20 \cdot \frac{1}{\sqrt{17}}$
   • Calculate $F$ from the second direction formula.

Using the equation: $F = \mu R$

4. Now, substitute the known values for $R$ and $F$ to find $\mu$:
   • From the earlier calculations, $R \approx 51.2 \text{ N}$ and $F \approx 13.4 \text{ N}$.
   • Thus, $\mu = \frac{F}{R} = \frac{13.4}{51.2} \approx 0.262$