In Fig. 8-46, a spring with k=170 N/mis at the top of a frictionless incline of angle$\mathbf{\theta }\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}$. The lower end of the incline is distance D = 1.00 mfrom the end of the spring, which is at its relaxed length. A 2.00 kgcanister is pushed against the spring until the spring is compressed 0.200 mand released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Spring constant, k = 170 N/m

Angle, $D=1.00\mathrm{m}$

Distance, D = 1.00 m

Mass, m = 2.00 kg

Spring compressed, x = 0.200 m

Acceleration due to gravity, $g=9.8\mathrm{m}/{\mathrm{s}}^2$

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

Using the principle of conservation of energy, this problem can be solved.

${\mathit{K}}_{\mathbf{i}}\mathbf{+}{\mathit{U}}_{\mathbf{i}}\mathbf{=}{\mathit{K}}_{\mathbf{f}}\mathbf{+}{\mathit{U}}_{\mathbf{f}}$

All heights h are measured from the lower end of the incline (which is our reference

position for computing gravitational potential energy ).

Our x-axis is along the incline, with +X being uphill (so spring compression corresponds to X>0 ) and its origin being at the relaxed end of the spring. The height that corresponds to the canister's initial position is given by

$h_1=\left(D+x\right)\sin \mathrm{\theta }$

Energy conservation leads to,

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ 0+mg\left(D+x\right)\sin \mathrm{\theta }+\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+mgD\sin \mathrm{\theta } \\ \frac{1}{2}m{v}_2^2=mg\left(D+x\right)\mathrm{sin\theta }-mgD\sin \mathrm{\theta }+\frac{1}{2}k{x}^2 \\ {v}_2^2=\frac{2mg\sin \mathrm{\theta }\left(\mathrm{D}+\mathrm{x}-1\right)}{m}+\frac{2}{2m}k{x}^2 \\ {v}_2=\sqrt{2g\sin \mathrm{\theta }\left(\mathrm{D}+\mathrm{x}-1\right)+\frac{{\mathrm{kx}}_2}{\mathrm{m}}} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{v}}_2=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin \left(37.0°\right)\left(1\mathrm{m}+0.200\mathrm{m}-1\right)+\frac{\left(170\mathrm{N}/\mathrm{m}\right){\left(0.200\mathrm{m}\right)}^2}{2.00\mathrm{kg}}} \\ =\sqrt{\left(19.6{\mathrm{m}}^2/{\mathrm{s}}^2\right)\left(0.602\right)\left(0.200\mathrm{m}\right)+\left(3.4\mathrm{N}.\mathrm{kg}/\mathrm{m}\right)} \\ =\sqrt{2.36{\mathrm{m}}^2/{\mathrm{s}}^2+3.4{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\sqrt{5.76{\mathrm{m}}^2/{\mathrm{s}}^2} \\ {\mathrm{v}}_2=2.40\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the canister at the instant the spring returns to its relaxed length is 2.40 m/s.

In this case, energy conservation leads to the following equation.

$$\begin{gathered} K_1+U_1=K_3+U_3 \\ 0+mg\left(D+x\right)\sin \theta +\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_3^2+0 \\ {v}_3^2=\frac{2mg\left(D+x\right)\sin \theta +k{x}^2}{m} \\ {v}_3=\sqrt{2g\left(D+x\right)\sin \theta +\frac{k{x}^2}{m}} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{v}}_3=\sqrt{2\mathrm{x}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1+0.200\right)\sin 37°+\frac{\left(170\mathrm{N}/\mathrm{m}\right){\left(0.200\mathrm{m}\right)}^2}{2.00\mathrm{kg}}} \\ =\sqrt{\left(19.6\mathrm{m}/{\mathrm{s}}^2\right)\left(1.200\right)\left(0.602\right)+\left(3.4\mathrm{N}.\mathrm{kg}/\mathrm{m}\right)} \\ =\sqrt{14.159{\mathrm{m}}^2/{\mathrm{s}}^2+3.4{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\sqrt{17.559{\mathrm{m}}^2/{\mathrm{s}}^2} \\ {\mathrm{v}}_3=4.19\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the canister when it reaches the lower end of the incline is 4.19 m/s.