In the diagram below, a small object of mass 2 kg is sliding at a constant velocity of 1,5 m s⁻¹ down a rough plane inclined at 7° to the horizontal surface - NSC Physical Sciences - Question 2 - 2017 - Paper 1

In the free-body diagram, the following forces should be represented:

• The weight of the object ($W$) acting downward, which is calculated as: $W = mg = 2 imes 9.8 ext{ N} = 19.6 ext{ N}$
• The normal force ($N$) acting perpendicular to the inclined plane.
• The frictional force ($f_k$) acting up the slope.

The components of weight parallel and perpendicular to the incline are:

• Perpendicular: W_{ot} = W imes ext{cos}(7^ ext{°})
• Parallel: $W_{ ext{parallel}} = W imes ext{sin}(7^ ext{°})$

The frictional force ($f_k$) can be calculated. Since the object is sliding at constant velocity, the frictional force equals the component of weight down the incline: $f_k = mg imes ext{sin}(7^ ext{°})$ Substituting values: $f_k = 2 imes 9.8 imes ext{sin}(7^ ext{°}) = (2)(9.8)(0.1219) ext{ N} \\ = 2.39 ext{ N}$

To find the coefficient of kinetic friction ($\, ext{μ}_k$), the relationship: $f_k = ext{μ}_k N$ Where $N = mg imes ext{cos}(7^ ext{°})$ can be utilized: $N = 2 imes 9.8 imes ext{cos}(7^ ext{°}) \\$

Thus, substituting in: $ext{μ}_k = \frac{f_k}{N} = \frac{2.39}{2 imes 9.8 imes ext{cos}(7^ ext{°})} = \frac{2.39}{(2)(9.8)(0.9945)} \\ ≈ 0.12$

Using the kinetic energy principle, where the initial kinetic energy ($K.E$) is converted to work done against friction: $K.E = \frac{1}{2}mv^2$ Setting into the work-energy principle: $W = f_k imes d$ Where $d$ is the distance travelled.

Equating: $\frac{1}{2}mv^2 = f_k imes d$ Solving for $d$: $d = \frac{\frac{1}{2}mv^2}{f_k}$ Substituting values: $d = \frac{\frac{1}{2}(2)(1.5^2)}{2.39} \\ = 0.957 ext{ m}$