Step 1: Understanding the Concept:
This problem involves analyzing the forces acting on a block on an inclined plane. To find the force required to just start moving the block upwards, we need to consider the forces that oppose this motion: the component of gravity acting down the incline and the force of kinetic friction (or static friction at the point of moving).
Step 2: Key Formula or Approach:
1. Draw a free-body diagram of the block on the inclined plane.
2. Resolve the gravitational force ($mg$) into components parallel and perpendicular to the incline.
3. Identify all forces acting on the block: applied force (F), gravitational component down the incline, normal force (N), and frictional force ($f$).
4. Apply Newton's First Law (condition for equilibrium or impending motion) in the directions parallel and perpendicular to the incline.
- Perpendicular to incline: $\sum F_{\perp} = 0$
- Parallel to incline: $\sum F_{\parallel} = 0$
5. The force of kinetic friction is given by $f = \mu N$.
Step 3: Detailed Explanation:
Let's analyze the forces:
- Gravitational Force ($mg$): Acts vertically downwards.
- Component perpendicular to the incline: $mg \cos\theta$.
- Component parallel to the incline (acting downwards): $mg \sin\theta$.
- Normal Force (N): Acts perpendicular to the incline, outwards. From equilibrium in the perpendicular direction, $N = mg \cos\theta$.
- Applied Force (F): Acts parallel to the incline, upwards (to move the block up).
- Frictional Force (f): Opposes the motion (or tendency of motion). Since we want to move the block up, friction acts down the incline. The maximum static friction (or kinetic friction if moving) is $f = \mu N = \mu mg \cos\theta$.
For the block to move up the plane at a constant velocity (or to be on the verge of moving up), the applied force F must balance all the forces acting down the incline.
\[ F = (\text{Gravitational component down the incline}) + (\text{Frictional force down the incline}) \]
\[ F = mg \sin\theta + f \]
Substitute the expression for the frictional force:
\[ F = mg \sin\theta + \mu mg \cos\theta \]
Step 4: Final Answer:
The force required to move the block up the inclined plane is $mg \sin\theta + \mu mg \cos\theta$. Therefore, option (B) is correct.