A wooden plank is inclined at and angle of 23 degrees above the horizontal. A 2.0 kg box attached to a rope is is pulled up the incline. The coefficient of kinetic friction between the box and the wooden plank is 0.15.

Question

A wooden plank is inclined at and angle of 23 degrees above the horizontal. A 2.0 kg box attached to a rope is is pulled up the incline. The coefficient of kinetic friction between the box and the wooden plank is 0.15.

What does the tension in the rope is the box moves up the incline at a constant velocity?

Answer 1

gravity down the plank: mg*sinTheta

friction down the plank: mg*mu*cosTheta

So if it is moving up at constant velocity, then tension= sum of friction+gravitydownplank

Answer 2

To find the tension in the rope as the box moves up the incline at a constant velocity, we need to consider the forces acting on the box.

First, we need to identify the forces acting on the box:
1. The weight of the box acting vertically downwards.
2. The normal force exerted by the plank acting perpendicular to the incline.
3. The frictional force acting parallel to the incline, opposing the motion.
4. The tension force exerted by the rope, directed up the incline.

Since the box is moving at a constant velocity, we know that the net force acting on the box is zero.

Now, let's analyze the forces in more detail:

1. The weight of the box (mg) can be resolved into two components: one perpendicular to the incline (mg*cosθ) and one parallel to the incline (mg*sinθ), where θ is the angle of inclination (23 degrees).

2. The normal force (N) exerted by the plank acts perpendicular to the incline. Since the box is not accelerating vertically, the normal force is equal in magnitude to the perpendicular component of the weight: N = mg*cosθ.

3. The frictional force (f) acts parallel to the incline and opposes the motion. The magnitude of the frictional force can be calculated using the equation: f = μ*N, where μ is the coefficient of kinetic friction. In this case, μ = 0.15, and N = mg*cosθ.

4. The tension force (T) exerted by the rope acts up the incline. Since the box is moving at a constant velocity, the tension force must be equal in magnitude to the parallel component of the weight (mg*sinθ) plus the frictional force (f). Therefore, T = mg*sinθ + f.

To calculate the tension in the rope, we substitute the values into the equation:
T = mg*sinθ + f
T = 2.0 kg * 9.8 m/s^2 * sin(23 degrees) + 0.15 * (2.0 kg * 9.8 m/s^2 * cos(23 degrees))

Evaluating these calculations will give us the numerical value of the tension in the rope.