Question

A 5 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by Fs = 45 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?

Answer 1

The work done is 315 J

Explanation:

The graph is missing: find it in attachment.

The work done by a variable force is given by

$W=\int F(x) dx$

where

F(x) is the magnitude of the force

x is the position

On a force vs position graph, the work done is equivalent to the area under the graph. Therefore, in order to find the work done as the block moves from x = 0 to x = 8.0 m, we have to calculate the area under the graph between these two points.

Each square corresponds (vertically) to 45 N, so the area of the first trapezium between x = 0 and x = 4.0 m is

$A_1 = \frac{(4+2)\cdot 2\cdot 45)}{2}=270$

Then we have to find the area of the triangle between x = 6.0 m and x = 8.0 m:

$A_2 = \frac{1}{2}(2)(45)=45$

Therefore, the total area (and the work done) is

$W=A_1+A_2=270+45=315 J$

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