A 400 N box is being pushed across a level floo... - JAMB Physics 2023 Question

Given:
- Weight of the box (\(W\)) = 400 N
- Applied force (\(P\)) = 100 N
- Angle (\(\theta\)) = 30°

To find the coefficient of kinetic friction (\(\mu_k\)), the forces acting along the horizontal direction are the frictional force (\(F_r\)) and the horizontal component of the applied force (\(P \cos 30^\circ\)).

The equation for equilibrium in the horizontal direction is:
\[ P \cos 30^\circ - F_r = ma \]

Since the box is moving at a constant speed, its acceleration is zero:
\[ P \cos 30^\circ - F_r = 0 \]

Now, we know that \(F_r = \mu R\), where \(R\) is the normal reaction. The normal reaction (\(R\)) is found by considering forces in the vertical direction:
\[ R - P \sin 30^\circ - W = 0 \]

Solving for \(R\):
\[ R = P \sin 30^\circ + W \]

Now, substitute \(R\) back into the equation for \(F_r\):
\[ P \cos 30^\circ - \mu R = 0 \]

\[ P \cos 30^\circ = \mu R \]

\[ \mu = \frac{P \cos 30^\circ}{R} \]

Substitute the given values:
\[ \mu = \frac{100 \cos 30^\circ}{450} \]

Simplify:
\[ \mu = \frac{100 \cdot \frac{\sqrt{3}}{2}}{450} \]

\[ \mu \approx 0.19 \]

Therefore, the correct option is 0.19