Question

5. The graph in Figure 10-16 shows the force

Answer 1

Answer: W = 0.8 J

Explanation:

At 0.12 m the force is 3.0 N

At 0.28 m the force is 7.0 N

The distance of stretch is 0.28 - 0.12 = 0.16 m

The AVERAGE force is (3.0 + 7.0) / 2 = 5.0 N

The work is

W =  Fd = 5.0(0.16) = 0.8 J

or

A formula guy might find the slope of the line

k =    7 N / 0.28 m = 25 N/m

PS = ½kx²

PS = ½(25)0.28² - ½(25)0.12² = 0.8 J

or

A charts gal might realize that the work is the area under the curve between the limits

W = A = (0.28 - 0.12)[3.0 + ½(7.0 - 3.0)] = 0.8 J

or

A math guy might find an equation for the line and integrate between the limits.

y = 25x

W = $\int\limits^a_b {25x} \, dx$  for b = 0.12 and a = 0.28

W = 12.5x² | for b = 0.12 and a = 0.28

W = 12.5(0.28)² - 12.5(0.12)²

W = 0.8 J