Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use `costheta=1-theta^2/2 and sintheta=theta` for small theta`.

Explanation

To find the tension in the string of the pendulum bob at the given instant, we can follow these steps:

### Step 1: Identify the forces acting on the bob
At the angle θ with the vertical, two forces act on the bob:
1. The gravitational force (mg) acting downward.
2. The tension (T) in the string acting along the string.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- A component along the direction of the tension: \( mg \cos \theta \)
- A component perpendicular to the tension: \( mg \sin \theta \)

### Step 3: Write the equation of motion
In the vertical direction, the forces must balance. Thus, the tension T can be expressed as:
\[ T = mg \cos \theta + \frac{mv^2}{r} \]
where:
- \( v \) is the speed of the bob (1.4 m/s)
- \( r \) is the radius of the circular path, which in this case is equal to the length of the pendulum (1 m).

### Step 4: Substitute the known values
Given:
- Mass \( m = 100 \text{ grams} = 0.1 \text{ kg} \)
- Gravitational acceleration \( g = 10 \text{ m/s}^2 \)
- Speed \( v = 1.4 \text{ m/s} \)
- Angle \( \theta = 0.20 \text{ rad} \)

Now, we can substitute these values into the equation.

### Step 5: Calculate \( mg \cos \theta \)
Using the approximation \( \cos \theta \approx 1 - \frac{\theta^2}{2} \):
\[ \cos(0.20) \approx 1 - \frac{(0.20)^2}{2} = 1 - \frac{0.04}{2} = 1 - 0.02 = 0.98 \]
Thus,
\[ mg \cos \theta = 0.1 \times 10 \times 0.98 = 0.98 \text{ N} \]

### Step 6: Calculate \( \frac{mv^2}{r} \)
\[ \frac{mv^2}{r} = \frac{0.1 \times (1.4)^2}{1} = 0.1 \times 1.96 = 0.196 \text{ N} \]

### Step 7: Combine the results to find T
Now substitute back into the tension equation:
\[ T = mg \cos \theta + \frac{mv^2}{r} \]
\[ T = 0.98 + 0.196 = 1.176 \text{ N} \]

### Step 8: Round to appropriate significant figures
Rounding to two decimal places, we get:
\[ T \approx 1.18 \text{ N} \]

### Final Result
The tension in the string at the instant when the bob is moving at 1.4 m/s and makes an angle of 0.20 radian with the vertical is approximately **1.18 N**.

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