91Ó°ÊÓ

The potential energy stored in a compressed or stretched spring is given by the formula:\[ U = \frac{1}{2} k x^2 \]where \( k \) is the spring constant, and \( x \) is the compression or extension.Here, \( k = 170 \; \mathrm{N/m} \) and \( x = 0.200 \; \mathrm{m} \).Substituting the values:\[ U = \frac{1}{2} \times 170 \times (0.200)^2 = 3.4 \; \mathrm{J} \]The potential energy stored in the spring is 3.4 J.

When the spring returns to its relaxed length, the potential energy converts into kinetic energy. At this point, there is no potential energy.By the conservation of energy:\[ \text{Initial Spring potential energy} = \text{Kinetic energy at relaxed length} \]\[ 3.4 = \frac{1}{2} m v^2 \]where \( m = 2.00 \; \mathrm{kg} \) and \( v \) is the speed to be determined.Re-arranging the equation for \( v \), we have:\[ v = \sqrt{\frac{2 \times 3.4}{2.00}} = \sqrt{3.4} \approx 1.84 \; \mathrm{m/s} \]The speed of the canister when the spring returns to its relaxed length is approximately 1.84 m/s.

Now, consider the motion down the incline after the spring has relaxed completely. The gravitational potential energy loss converts to kinetic energy as the canister moves down the incline.The height \( h \) climbed/descended on the incline:\[ h = D \cdot \sin(\theta) = 1.00 \cdot \sin(37.0^{\circ}) \]Use \( \sin(37.0^{\circ}) \approx 0.602 \)\[ h = 1.00 \cdot 0.602 = 0.602 \; \mathrm{m} \]Gravitational potential energy loss:\[ \Delta U = mgh = 2.00 \cdot 9.8 \cdot 0.602 \approx 11.8 \; \mathrm{J} \]This energy also becomes kinetic energy:\[ \Delta U = \frac{1}{2} mv_{\text{bottom}}^2 - \frac{1}{2} mv_{\text{relaxed}}^2 \]Substitute the known values:\[ 11.8 = \frac{1}{2} \times 2.00 \times v_{\text{bottom}}^2 - 3.4 \]\[ v_{\text{bottom}}^2 = \frac{11.8 + 3.4}{1} = 15.2 \]\[ v_{\text{bottom}} = \sqrt{15.2} \approx 3.90 \; \mathrm{m/s} \]The speed of the canister at the lower end of the incline is approximately 3.90 m/s.