A 0.075-kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow ?

Options

A. 39m/s
B. 28m/s
C. 82m/s
D. 66m/s

Answer

To solve the problem step by step, we will use the concepts of Newton's second law of motion and the equations of motion.

### Step 1: Identify the given data
- Mass of the arrow (m) = 0.075 kg
- Average force exerted by the bowstring (F) = 65 N
- Distance over which the force is applied (d) = 0.90 m

### Step 2: Calculate the acceleration of the arrow
Using Newton's second law, we know that:
\[ F = m \cdot a \]
Where:
- \( F \) is the force,
- \( m \) is the mass,
- \( a \) is the acceleration.

Rearranging the formula to find acceleration:
\[ a = \frac{F}{m} \]

Substituting the values:
\[ a = \frac{65 \, \text{N}}{0.075 \, \text{kg}} \]
\[ a = 866.67 \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find the initial velocity
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (which is 0 when the arrow comes to rest),
- \( u \) is the initial velocity (which we want to find),
- \( a \) is the acceleration (which will be negative since the arrow is coming to rest),
- \( s \) is the distance (0.90 m).

Since the arrow is fired horizontally and we want to find the speed at which it leaves the bow, we can consider the acceleration as negative:
\[ v^2 = u^2 + 2(-a)s \]

Rearranging the equation to find \( u \):
\[ u^2 = v^2 - 2as \]

Substituting the values:
\[ u^2 = 0 - 2(-866.67)(0.90) \]
\[ u^2 = 2 \times 866.67 \times 0.90 \]
\[ u^2 = 1560 \]
\[ u = \sqrt{1560} \]
\[ u \approx 39.49 \, \text{m/s} \]

### Step 4: Conclusion
The speed at which the arrow leaves the bow is approximately **39.49 m/s**.

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