We can approach this problem by analyzing the forces and applying Newton's second law, both for linear and rotational motion.

Step 1: Free-body diagram of the block

For the mass $m = 14.9 \, \text{kg}$ on the inclined plane, the forces acting on it are:

• Gravitational force: $mg$
• Tension in the rope: $T$
• Normal force from the inclined plane

The gravitational force can be broken down into two components:

• $mg \sin(\theta)$: parallel to the incline, causing the acceleration
• $mg \cos(\theta)$: perpendicular to the incline, balanced by the normal force

Step 2: Newton's second law (linear motion)

Since the block is accelerating down the incline with acceleration $a = 2.02 \, \text{m/s}^2$, we can apply Newton's second law in the direction of motion:

$mg \sin(\theta) - T = ma$

Solving for the tension $T$:

$T = mg \sin(\theta) - ma$

We know:

• $m = 14.9 \, \text{kg}$
• $g = 9.8 \, \text{m/s}^2$
• $\theta = 33.7^\circ$
• $a = 2.02 \, \text{m/s}^2$

Substitute these values:

$T = (14.9)(9.8) \sin(33.7^\circ) - (14.9)(2.02)$

First, calculate each term: $mg \sin(33.7^\circ) = 14.9 \times 9.8 \times 0.5543 = 81.07 \, \text{N}$ $ma = 14.9 \times 2.02 = 30.10 \, \text{N}$

Now, subtract: $T = 81.07 - 30.10 = 50.97 \, \text{N}$

Thus, the tension in the rope is approximately 50.97 N.

Would you like further explanation, or do you have any questions?

Here are some related questions to explore:

1. How would the tension change if the incline angle increased?
2. What would be the effect of adding friction to the incline?
3. How is the rotational inertia of the wheel involved in this scenario?
4. What if the wheel had friction at its axle—how would this affect the tension?
5. Could you determine the torque acting on the wheel from the given data?

Tip: Always check whether forces should be broken into components based on their direction relative to the motion.