The expression for the magnitude of the frictional force on the block if it is moving at constant speed is μk (Fsinθ + mg). Option D.
When an object is moving on a surface, the frictional force is known as kinetic friction. The magnitude of the kinetic frictional force is given by the equation:
f_k = μ_k * N
where:
- f_k is the kinetic frictional force,
- μ_k is the coefficient of kinetic friction, and
- N is the normal force.
The normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular (or "normal") to the surface. The normal force can be affected by other forces acting on the object, such as an external force applied at an angle.
In this problem, the block is being pulled by a force F at an angle θ. This force has two components:
- Fcosθ acting horizontally, and
The vertical component of the force Fsinθ affects the normal force. When the block is pulled upwards, it partially offsets the weight of the block, decreasing the normal force. Conversely, if the block were pulled downwards, it would increase the normal force.
So, in this case, the normal force N on the block is the weight of the block mg plus the vertical component of the pulling force Fsinθ. Therefore:
N = mg + Fsinθ
Substituting this into the equation for kinetic friction, we get:
μk (Fsinθ + mg)