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An *m* = 14.9kg mass is attached to a cord that is wrapped around a wheel of radius *r* = 12.8cm (see the figure below). The acceleration of the mass down the frictionless incline is measured to be *a* = 2.02m/s^2. Assuming the axle of the wheel to be frictionless, and the angle to be θ = 33.7deg determine the tension in the rope.

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.

An m = 14.9kg mass is attached to a cord that is wrapped around a wheel of radius r = 12.8cm. The acceleration of the mass down the frictionless incline is measured to be a = 2.02m/s^2. Assuming the axle of the wheel to be frictionless, and the angle to be θ = 33.7°, determine the tension in the rope.

Determine the tension in a rope attached to a mass on a frictionless incline using Newton's second law. The problem involves a 14.9kg mass, 2.02m/s^2 acceleration, and an incline angle of 33.7°, with a wheel radius of 12.8cm. Suitable for physics students in Grades 10-12.

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