Final answer:
When a 2-kilogram block slides down a 30° rough incline with constant velocity, the magnitude of the frictional force along the plane is approximately 10 N. The correct answer is option d.
Step-by-step explanation:
When a 2-kilogram block slides down a 30° rough incline with constant velocity, the frictional force along the plane is equal in magnitude but opposite in direction to the component of the weight of the block parallel to the incline.
Since the block is sliding with constant velocity, the net force on the block in the direction of motion is zero.
The magnitude of the component of the weight parallel to the incline can be calculated using the formula:
Force = mass * acceleration
Since the block is sliding with constant velocity, the acceleration is zero. Therefore, the magnitude of the frictional force along the plane is equal to the magnitude of the weight of the block parallel to the incline. Using trigonometry, we can calculate the magnitude of the weight of the block parallel to the incline as:
Weight parallel = weight * sin(angle of incline)
Substituting the values, we get:
Weight parallel = 2 kg * 10 m/s² * sin(30°)
Weight parallel = 2 kg * 10 m/s² * 0.5
Weight parallel = 10 N
Therefore, the magnitude of the frictional force along the plane is approximately 10 N.