Question 8.6: A Block–Spring System A block of mass 1.6 kg is attached to ......

(A) Conceptualize This situation has been discussed before, and it is easy to visualize the block being pushed to the right by the spring and moving off with some speed.

Categorize We identify the system as the block and model the block as a nonisolated system.

Analyze In this situation, the block starts with v_{i}=0 at x_{i}=-2.0 \mathrm{~cm}, and we want to find v_{f} at x_{f}=0.

Use Equation 7.11

W_s=\int \overrightarrow{\mathbf{F}}_s \cdot d \overrightarrow{\mathbf{r}}=\int_{x_i}^{x_f}(-k x \hat{\mathbf{i}}) \cdot(d x \hat{\mathbf{i}})=\int_{-x_{\max }}^0(-k x) d x=\frac{1}{2} k x_{\max }^2       (7.11)

to find the work done by the spring on the system with x_{\max }=x_{i} :

\sum W_{\text {other forces }}=W_{s}=\frac{1}{2} k x_{\max }^{2}

Work is done on the block, and its speed changes. The conservation of energy equation, Equation 8.2,

\Delta K+\Delta U+\Delta E_{\mathrm{int}}=W+Q+\;T_{\mathrm{MW}}+\;T_{\mathrm{MT}}+\;T_{\mathrm{ET}}+\;T_{\mathrm{ER}}         (8.2)

reduces to the work-kinetic energy theorem. Use that theorem to find the speed at x=0 :

\begin{aligned}& W_{s}=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\&v_{f}=\sqrt{v_{i}^{2}+\frac{2}{m} W_{s}}\end{aligned}

Substitute numerical values:

v_{f}=\sqrt{0+\frac{2}{1.6 \mathrm{~kg}}\left[\frac{1}{2}(1000 \mathrm{~N} / \mathrm{m})(0.020 \mathrm{~m})^{2}\right]}=0.50 \mathrm{~m} / \mathrm{s}

Finalize Although this problem could have been solved in Chapter 7, it is presented here to provide contrast with the following part (\mathrm{B}), which requires the techniques of this chapter.

(B) Conceptualize The correct answer must be less than that found in part (A) because the friction force retards the motion.

Categorize We identify the system as the block and the surface. The system is nonisolated because of the work done by the spring, and there is a nonconservative force acting: the friction between the block and the surface.

Analyze Write Equation 8.14:

K_{f}=K_{i}-f_{k}d+\sum W_{\text{other forces}}         (8.14)

(1) K_{f}=K_{i}-f_{k} d+W_{s}

Substitute numerical values:

K_{f}=0-(4.0 \mathrm{~N})(0.020 \mathrm{~m})+\frac{1}{2}(1000 \mathrm{~N} / \mathrm{m})(0.020 \mathrm{~m})^{2}=0.12 \mathrm{~J}

Write the definition of kinetic energy:

K_{f}=\frac{1}{2} m v_{f}^{2}

Solve for v_{f} and substitute numerical values:

v_{f}=\sqrt{\frac{2 K_{f}}{m}}=\sqrt{\frac{2(0.12 \mathrm{~J})}{1.6 \mathrm{~kg}}}=0.39 \mathrm{~m} / \mathrm{s}

Finalize As expected, this value is less than the 0.50 \mathrm{~m} / \mathrm{s} found in part (\mathrm{A}).