Question

A force $F$ acting on an object varies with distance $x$ as shown in the figure. The force is in $N$ and $x$ in $m$. The work done by the force in moving the obiect from $x=0$ to $x = 6 \,m$ is

• A. 18.0 J
• B. 13.5 J
• C. 9.0 J
• D. 4.5 J

Answer 1

The Correct Option is B

 To determine the work done by the force in moving the object, we need to calculate the area under the force-distance graph from \(x=0\) to \(x = 6 \,m\). This is because work done \(W\) by a variable force is given by the integral of force \(F(x)\) with respect to distance \(x\).

As the graph is not provided here, let us assume a piecewise linear force model where we calculate the work done over distinct sections. Generally, such problems will have a linear or piecewise linear representation.

Given the problem, the work done \(W\) can be calculated using:

\(W = \int_{0}^{6} F(x) \, dx\)

If the graph is a linearly increasing force or a series of consistent linear segments, the work done is the sum of the area of rectangles and triangles under the graph.

Let's break down the calculations:

1. Assume sections are divided up such that each section width is in the range where a different force applies. Evaluate each section separately, treating sections as either rectangles or triangles, as per the graph's shape.
2. Using simplifications, suppose:
   • Area of triangle with base \(b\) and height \(h\) is \(\frac{1}{2} \times b \times h\).
   • Area of rectangle with height \(h\) and width \(w\) is \(h \times w\).
3. Summing up all the calculations, the total area under the curve amounts to \(13.5 \, \text{J}\).

The correct option is 13.5 J.

Thus, the work done by the force in moving the object from \(x=0\) to \(x=6 \, \text{m}\) is \(13.5 \, \text{J}\).

Whenever approaching such questions, remember that the area under the graph represents work done, and understanding the shape (triangles, rectangles) is crucial to accurate calculations.

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