Chapter 14, Problem 102A

(a)

To determine

To graph of force applied to a spring versus spring length and plot the force on y-axis from a given table of force and length of spring extensions.

Explanation of Solution

Introduction:

Given table is,

Force magnitude, F(N)	Extension, x(m)
2.5	0.12
5.0	0.26
7.5	0.35
10.0	0.50
12.5	0.60
15.0	0.71

The graph of the given table is plotted as force(F) versus spring length(x),

  

Conclusion:

Hence, the required answers are given above.

(b)

To determine

To find: the spring constant from the graph.

Explanation of Solution

Introduction:

Given table is,

Force magnitude, F(N)	Extension, x(m)
2.5	0.12
5.0	0.26
7.5	0.35
10.0	0.50
12.5	0.60
15.0	0.71

The spring constant can be determined from the slope of the graph plotted above,

  $k=slope=\frac{\Delta F}{\Delta x}=\frac{F_{final}-F_{initial}}{x_{final}-x_{initial}}$ ………(1)

Substituting $15$ N in F_final, $2.5\text{ N}$ in F_initial, $\text{0}\text{.71 m}$ in x_finaland $\text{0}\text{.12 m}$ in x_initial in equation (1)

  $k=\frac{15\text{ N}-2.5\text{ N}}{0.71\text{ m}-0.12\text{ m}}=21\text{ N/m}$

Conclusion:

Hence, the required potential energy is $21\text{N/m}$ .

(c)

To determine

To find: the elastic potential energy stored in the spring when stretched by 5.0m using the graph.

Explanation of Solution

Introduction:

Given table is,

Force magnitude, F(N)	Extension, x(m)
2.5	0.12
5.0	0.26
7.5	0.35
10.0	0.50
12.5	0.60
15.0	0.71

The potential energy can be determined from the area under the graph.

  $P.E=area=\frac{1}{2}bh$ ……….(2)

Here, b is the breadth and h is the height.

Substituting 0.50 m in b and 10 N in h

  $P.E=\frac{1}{2}(0.50)(10)=2.5\text{ J}$

Conclusion:

Hence, the required potential energy is $2.5\text{J}$ .