To determine the net force acting on the barge, we need to consider all the forces acting on it and their directions.
- Buoyant force (F_buoy): 25,000 N (upward)
- Pulling force (F_pull): 350 N (rightward)
- Friction force (F_friction): 100 N (leftward)
- 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.