Question

The following free body diagram represents the forces acting on a barge being towed by tugboat into a
harbor.

Fbuoy = 25,000 N

Fpull = 350 N

Ffriction = 100 N

Fw = 25,000 N

What is the net force acting on the barge?

250 N to the left

100 N to the right

300 N to the right

350 N to the left

If the mass of the barge is 12,000 kg, what is the acceleration it experiences? (1 point)

0.981 m/s^2 to the right

0.021 m/s^2 to the left

3.91 m/s^2 to the left

0.067 m/s^2 to the right

Answer 1

To determine the net force acting on the barge, we need to consider all the forces acting on it and their directions.

1. **Buoyant force (F_buoy)**: 25,000 N (upward)
2. **Pulling force (F_pull)**: 350 N (rightward)
3. **Friction force (F_friction)**: 100 N (leftward)
4. **Weight of the barge (Fw)**: 25,000 N (downward)

We can find the net horizontal force first:

- The net horizontal force (F_net_horizontal) is given by the pulling force minus the friction force:
\[
F_{net\_horizontal} = F_{pull} - F_{friction} = 350 \, \text{N} - 100 \, \text{N} = 250 \, \text{N} \, \text{to the right}
\]

Since vertical forces (buoyant and weight) do not affect horizontal motion, we focus on the horizontal forces.

Now, the net force acting on the barge is:
- **250 N to the right**.

Therefore, the answer to the first question is:
**250 N to the right**

---

Next, for the acceleration of the barge, we can use Newton's second law:
\[
F_{net} = m \times a
\]
Where:
- \( F_{net} \) is the net force (250 N to the right)
- \( m \) is the mass of the barge (12,000 kg)
- \( a \) is the acceleration

Reorganizing the equation to solve for acceleration:
\[
a = \frac{F_{net}}{m}
\]
Substituting in the values:
\[
a = \frac{250 \, \text{N}}{12,000 \, \text{kg}} \approx 0.02083 \, \text{m/s}^2
\]
- This round to 0.021 m/s².

Since the force is to the right, the acceleration will also be to the right.

Therefore, the answer to the second question is:
**0.021 m/s^2 to the right**.