The net work done on a cart moving on a level track, as shown by a force-displacement graph, is 10 J. This results in a net increase in kinetic energy of the cart as it travels from 0 m to 4 m in the positive x-direction.
The net work done on an object is equal to the change in its kinetic energy according to the work-energy theorem. The net work (\(W_{\text{net}}\)) can be calculated as the area under the force vs. displacement graph.
From the graph, we can divide the displacement into three segments and calculate the area under the force-displacement curve for each segment:
1. The first segment: \(0 \, \text{m} \leq x \leq 2 \, \text{m}\) with a constant force of \(10 \, \text{N}\).
2. The second segment: \(2 \, \text{m} \leq x \leq 4 \, \text{m}\) with a constant force of \(-5 \, \text{N}\).
For the first segment, the work done (\(W_1\)) is the area of a rectangle:
\[ W_1 = F \cdot d = 10 \, \text{N} \cdot 2 \, \text{m} = 20 \, \text{J} \]
For the second segment, the work done (\(W_2\)) is the area of a rectangle:
\[ W_2 = F \cdot d = -5 \, \text{N} \cdot 2 \, \text{m} = -10 \, \text{J} \]
The total work done (\(W_{\text{net}}\)) is the sum of \(W_1\) and \(W_2\):
\[ W_{\text{net}} = W_1 + W_2 = 20 \, \text{J} + (-10 \, \text{J}) = 10 \, \text{J} \]
The net change in kinetic energy (\(\Delta KE\)) is equal to the net work done:
\[ \Delta KE = W_{\text{net}} = 10 \, \text{J} \]
Therefore, the correct answer is B. An increase of 10 J.