Question

work done by varying force example 14. the force acting on a particle varies as shown in figure p7.14. find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 to x = 10.0 m. figure p7.14

Answer:

To solve the work done by a varying force, we use the concept that the work done \( W \) by a force \( F(x) \) as a particle moves from \( x = x_1 \) to \( x = x_2 \) is the area under the \( F(x) \)-versus-\( x \) graph between those two points. Let's analyze each part:

Part (a): From \( x = 0 \) to \( x = 8.00 \, \text{m} \)

The graph from \( x = 0 \) to \( x = 8.00 \, \text{m} \) is a triangle. For a triangle, the area (work) is given by:

$$ W = \frac{1}{2} \times \text{base} \times \text{height} $$

• Base: \( 8.00 \, \text{m} \) (from \( x = 0 \) to \( x = 8 \))
• Height: The maximum force at the peak. From the graph, at \( x = 4 \, \text{m} \) (midpoint), the force is \( 6 \, \text{N} \)? Wait, no—wait, the graph starts at \( x = 0 \), \( F = 0 \)? Wait, no, looking at the graph: At \( x = 0 \), \( F = 0 \)? Wait, no, the leftmost point is at \( x = 0 \), \( F = 0 \)? Wait, the graph is a triangle? Wait, no, the graph shown: At \( x = 0 \), \( F = 0 \)? Wait, no, the first vertex is at \( x = 0 \), \( F = 0 \)? Wait, no, the graph has a vertex at \( x = 0 \), \( F = 0 \)? Wait, no, looking at the grid: The first point is at \( x = 0 \), \( F = 0 \)? Wait, no, the red line starts at \( x = 0 \), \( F = 0 \)? Wait, no, the graph is a triangle with vertices at \( (0, 0) \), \( (4, 6) \), and \( (8, 0) \)? Wait, no, the user’s graph: Let's re-examine. The \( x \)-axis is in meters, \( F_x \) in Newtons. The red line: At \( x = 0 \), \( F = 0 \); at \( x = 4 \, \text{m} \), \( F = 6 \, \text{N} \); at \( x = 8 \, \text{m} \), \( F = 0 \). So it’s a triangle with base \( 8.00 \, \text{m} \) and height \( 6 \, \text{N} \).

Thus:

$$ W = \frac{1}{2} \times 8.00 \, \text{m} \times 6 \, \text{N} = 24 \, \text{J} $$

Part (b): From \( x = 8.00 \, \text{m} \) to \( x = 10.0 \, \text{m} \)

From \( x = 8.00 \, \text{m} \) to \( x = 10.0 \, \text{m} \), the graph is a triangle below the \( x \)-axis (negative force, meaning work is negative).

• Base: \( 10.0 - 8.00 = 2.00 \, \text{m} \)
• Height: The magnitude of the force at \( x = 10 \, \text{m} \). From the graph, at \( x = 10 \, \text{m} \), \( F = -2 \, \text{N} \)? Wait, no, the vertex at \( x = 10 \, \text{m} \) has \( F = -2 \, \text{N} \)? Wait, the graph continues from \( x = 8 \) to \( x = 10 \), forming a smaller triangle below the \( x \)-axis.

For a triangle below the \( x \)-axis, the area (work) is negative (since force and displacement are in opposite directions).

• Base: \( 2.00 \, \text{m} \) (from \( x = 8 \) to \( x = 10 \))
• Height: The magnitude of the force at the peak (trough) is \( 2 \, \text{N} \) (since at \( x = 9 \, \text{m} \), \( F = -1 \, \text{N} \)? Wait, no, the vertex at \( x = 10 \, \text{m} \) is at \( F = -2 \, \text{N} \)? Wait, the graph’s vertex at \( x = 10 \, \text{m} \) is \( F = -2 \, \text{N} \). So the triangle from \( x = 8 \) to \( x = 10 \) has:
• Base: \( 2.00 \, \text{m} \)
• Height: \( 2 \, \text{N} \) (magnitude, but since it’s below the axis, work is negative).

Area (work):

$$ W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.00 \, \text{m} \times 2 \, \text{N} = 2 \, \text{J} $$

But since the force is negative (opposite to displacement), the work is \( -2 \, \text{J} \)? Wait, no—wait, displacement is from \( x = 8 \) to \( x = 10 \) (positive direction), and force is negative (opposite), so work is negative. Wait, but let's check the graph again. Wait, the vertex at \( x = 10 \, \text{m} \) is \( F = -2 \, \text{N} \)? Wait, the grid: Each square is 1 unit. At \( x = 10 \, \text{m} \), \( F = -2 \, \text{N} \). So the triangle from \( x = 8 \) to \( x = 10 \) has base \( 2 \, \text{m} \) and height \( 2 \, \text{N} \) (magnitude), but since it’s below the \( x \)-axis, the area (work) is negative:

$$ W = -\frac{1}{2} \times 2.00 \, \text{m} \times 2 \, \text{N} = -2 \, \text{J} $$

Part (c): From \( x = 0 \) to \( x = 10.0 \, \text{m} \)

This is the total work: work from \( 0 \) to \( 8 \) (part a) plus work from \( 8 \) to \( 10 \) (part b).

From part (a): \( W_{0 \to 8} = 24 \, \text{J} \)
From part (b): \( W_{8 \to 10} = -2 \, \text{J} \) (wait, no—wait, maybe I messed up part (b). Wait, let's re-express the graph correctly.

Wait, maybe the graph is:

• From \( x = 0 \) to \( x = 4 \, \text{m} \): force increases from \( 0 \) to \( 6 \, \text{N} \) (triangle)
• From \( x = 4 \) to \( x = 8 \, \text{m} \): force decreases from \( 6 \, \text{N} \) to \( 0 \) (triangle)
• From \( x = 8 \) to \( x = 10 \, \text{m} \): force decreases from \( 0 \) to \( -2 \, \text{N} \) (triangle)

Wait, no, the graph shown has a vertex at \( x = 10 \, \text{m} \), \( F = -2 \, \text{N} \). Let's re-express the areas:

Part (a): \( 0 \to 8 \, \text{m} \)

The shape from \( 0 \) to \( 8 \) is a triangle with base \( 8 \, \text{m} \) and height \( 6 \, \text{N} \) (at \( x = 4 \, \text{m} \), \( F = 6 \, \text{N} \)). Thus:

$$ W_{0 \to 8} = \frac{1}{2} \times 8 \, \text{m} \times 6 \, \text{N} = 24 \, \text{J} $$

Part (b): \( 8 \to 10 \, \text{m} \)

The shape from \( 8 \) to \( 10 \) is a triangle with base \( 2 \, \text{m} \) (10 - 8) and height \( 2 \, \text{N} \) (magnitude, but force is negative, so work is negative). Thus:

$$ W_{8 \to 10} = \frac{1}{2} \times 2 \, \text{m} \times (-2 \, \text{N}) = -2 \, \text{J} $$

Part (c): \( 0 \to 10 \, \text{m} \)

Total work is the sum of part (a) and part (b):

$$ W_{0 \to 10} = W_{0 \to 8} + W_{8 \to 10} = 24 \, \text{J} + (-2 \, \text{J}) = 22 \, \text{J} $$

Wait, but let's verify the graph again. Maybe the height at \( x = 4 \, \text{m} \) is \( 6 \, \text{N} \), so the first triangle (0–8 m) has area \( \frac{1}{2} \times 8 \times 6 = 24 \, \text{J} \). The second triangle (8–10 m) has base \( 2 \, \text{m} \) and height \( 2 \, \text{N} \) (but negative), so area \( \frac{1}{2} \times 2 \times (-2) = -2 \, \text{J} \). Thus total work from 0–10 m is \( 24 - 2 = 22 \, \text{J} \).

Final Answers:

(a) \( \boldsymbol{24 \, \text{J}} \)
(b) \( \boldsymbol{-2 \, \text{J}} \)
(c) \( \boldsymbol{22 \, \text{J}} \)

(Note: If the graph’s peak force is different, adjust accordingly. But based on the grid, this is the best interpretation.)