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The force exerted on a 8.0-kg block is shown in the figure as a function of time. How far does the block travel in the first 10.4 s? Assume that the motion is one dimensional and that the velocity of the block at time t= 0 s is 0 m/s.

To calculate the distance traveled by the 8.0-kg block in the first 10.4 seconds based on the force-time graph, we can follow these steps:

1. **Find the acceleration as a function of time**:  
   Use Newton's second law, \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. For this block:
   \[
   a = \frac{F}{m} = \frac{F}{8.0}
   \]

2. **Find the velocity as a function of time**:  
   The velocity is the integral of acceleration over time:
   \[
   v(t) = v_0 + \int_0^t a(t) \, dt
   \]
   Since the initial velocity is \( v_0 = 0 \, \text{m/s} \), the velocity at any time can be found by integrating the acceleration.

3. **Find the displacement**:  
   The displacement is the integral of velocity over time:
   \[
   x(t) = \int_0^t v(t) \, dt
   \]

### Step-by-step process:

We will break down the time intervals based on the force changes in the graph:
- From 0 to 2 seconds: Force = 4 N
- From 2 to 4 seconds: Force = 8 N
- From 4 to 6 seconds: Force = 6 N
- From 6 to 9 seconds: Force = 4 N
- From 9 to 10.4 seconds: Force = 6 N

For each interval, we will calculate the acceleration, velocity, and displacement.

### Step 1: Calculate accelerations for each interval.
- From \( 0 \leq t < 2 \) s: \( a = \frac{4}{8} = 0.5 \, \text{m/s}^2 \)
- From \( 2 \leq t < 4 \) s: \( a = \frac{8}{8} = 1.0 \, \text{m/s}^2 \)
- From \( 4 \leq t < 6 \) s: \( a = \frac{6}{8} = 0.75 \, \text{m/s}^2 \)
- From \( 6 \leq t < 9 \) s: \( a = \frac{4}{8} = 0.5 \, \text{m/s}^2 \)
- From \( 9 \leq t < 10.4 \) s: \( a = \frac{6}{8} = 0.75 \, \text{m/s}^2 \)

Let me now calculate the total displacement by integrating velocity over these intervals.The block travels a distance of approximately **37.39 meters** in the first 10.4 seconds, and its velocity at the end of this time period is about **7.05 m/s**.

Would you like further details on the calculations or the kinematic process?

Here are some related questions:
1. How would the result change if the initial velocity was not zero?
2. What would be the distance traveled if the force was constant?
3. How can the work done on the block be calculated from the force-time graph?
4. What is the average acceleration over the entire time period?
5. How would friction affect the block's motion in this scenario?

**Tip:** Breaking complex problems into smaller intervals, like we did with time, often simplifies integration in physics problems.

Calculate the distance traveled by an 8.0-kg block in the first 10.4 seconds using Newton's second law and integration. The problem involves force as a piecewise function over time and covers kinematic concepts such as velocity and displacement.

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