A lifeboat slides down a straight ramp inclined at an angle of 15° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 1

The forces acting on the lifeboat are:

• The gravitational force down the ramp: $F_g = mg \sin(\theta)$
• The normal reaction: $R = mg \cos(\theta)$

Substituting the mass $m = 800 \, \text{kg}$ and $\theta = 15^{\circ}$:

$F_g = 800 \times 9.8 \times \sin(15^{\circ})$

$R = 800 \times 9.8 \times \cos(15^{\circ})$

By applying Newton's second law, we set up the equation:

$F_g - f = ma$

where the friction force is given by $f = \mu R$. Thus,

$800g \sin(15^{\circ}) - \mu (800g \cos(15^{\circ})) = 800 \times 1.5876$

Solving for the coefficient of friction $\mu$:

$\mu = \frac{800g \sin(15^{\circ}) - 800 \times 1.5876}{800g \cos(15^{\circ})}$

Simplifying the equation:

1. Substitute $g \approx 9.8 \, \text{m/s}^2$:

   $\mu = \frac{(800 \times 9.8 \times \sin(15^{\circ})) - (800 \times 1.5876)}{800 \times 9.8 \times \cos(15^{\circ})}$

2. After calculating the values:

   The resulting coefficient of friction $\mu$ will approximate to: $0.1$, yielding valid options $\mu \/ 0.1, 0.10, 0.100$.