Question

determine the work done in the following graphs?
distance (m)

Explanation:

Step1: Identify the graph type

The graph is a Force - Distance graph (assuming the vertical axis is Force, let's say \( F \) and horizontal is Distance \( d \)). For a \( F - d \) graph, work done \( W=\int F\mathrm{d}d \), and if \( F \) is constant, \( W = F\times d \).

Step2: Determine values of F and d

From the graph, let's assume the vertical axis (Force) has a value, say \( F = 15\space N \) (since it's between 10 and 20, mid - point or as per grid, let's check distance: the horizontal length (distance) \( d \): from the graph, the distance covered (horizontal axis) – let's see the grid. If we assume each grid square on x - axis (distance) is 1 unit, and the length of the horizontal line is 4 units (from x = 0 to x = 4, since 5 is marked, but the line ends at x = 4 maybe? Wait, the graph's horizontal line: let's re - evaluate. Wait, the vertical axis: the force is constant. Let's suppose the force \( F = 15\space N \) (as it's a horizontal line between 10 and 20, so maybe 15 N). The distance \( d \): let's count the number of units on the x - axis. If the x - axis (distance) has a scale where from 0 to 4 (since the dot is at x = 4 maybe? Wait, the graph shows distance (m) with 5 marked. Wait, maybe the horizontal line is from \( d = 0 \) to \( d = 4\space m \) (assuming the grid squares: each square on x is 1 m). And force \( F = 15\space N \).

Step3: Calculate work done

Work done \( W=F\times d \). If \( F = 15\space N \) and \( d = 4\space m \), then \( W=15\times4 = 60\space J \). Wait, maybe the force is 15 N (since the vertical line is between 10 and 20, so 15 N) and distance is 4 m (from x = 0 to x = 4). Alternatively, maybe the force is 15 N and distance is 4 meters. Wait, let's check the graph again. The horizontal line (force) is constant. Let's assume the force \( F = 15\space N \) and distance \( d = 4\space m \) (the length of the horizontal segment). Then \( W=F\times d=15\times4 = 60\space J \).

Wait, maybe the force is 15 N and distance is 4 m. Let's confirm: in a force - distance graph, work done is the area under the force - distance curve. Since it's a rectangle (constant force, constant distance), area = length × width. So if force is \( F = 15\space N \) (height of the rectangle) and distance \( d = 4\space m \) (width of the rectangle), then area (work) is \( 15\times4 = 60\space J \).

Answer:

The work done is \( \boldsymbol{60\space J} \) (assuming \( F = 15\space N \) and \( d = 4\space m \); the values may vary slightly based on exact grid interpretation, but the method is \( W = F\times d \) for constant force in \( F - d \) graph).