In this question use $g = 9.8 \, \text{m s}^{-2}$ A rough wooden ramp is 10 metres long and is inclined at an angle of $25^{\circ}$ above the horizontal - AQA - A-Level Maths Pure - Question 19 - 2022 - Paper 2

To find the coefficient of friction, we begin by analyzing the forces acting on the crate. The weight of the crate can be resolved into two components: one parallel to the ramp and one perpendicular to the ramp.

1. Weight Resolution:

   • The weight of the crate is given by: $W = mg = 20 \, \text{kg} \times 9.8 \, \text{m s}^{-2} = 196 \, \text{N}$
   • The component of the weight parallel to the slope is: $W_{\parallel} = W \sin(25^{\circ}) = 196 \sin(25^{\circ})$(approx. $82.4 \, \text{N}$)
   • The component of the weight perpendicular to the slope is: $W_{\perpendicular} = W \cos(25^{\circ}) = 196 \cos(25^{\circ})$ (approx. $176.0 \, \text{N}$)

2. Applying Newton's Second Law:

   • The net force equation in the direction along the ramp is: $T - W_{\parallel} - f = ma$
   • Where:
     • $T$ is the tension in the rope ($230 \, \text{N}$)
     • $f$ is the frictional force, which can be expressed as: $f = \mu R = \mu (mg \cos(25^{\circ})) = \mu (196 \cos(25^{\circ}))$
   • Substituting the values into the equation yields: $230 - 82.4 - \mu (176.0) = 20 \times 1.2$
   • Simplifying gives: $230 - 82.4 - 176.0 \mu = 24$

3. Solving for the Coefficient of Friction:

   • Rearranging the equation: $176.0 \mu = 230 - 82.4 - 24$ $176.0 \mu = 123.6$ $\mu = \frac{123.6}{176.0} \\approx 0.70$