In solving the curved path problem, we aim to determine the minimum speed required for a block to ascend a hill of height 20 meters. This scenario emphasizes the importance of energy conservation, as traditional methods involving forces or kinematics are inadequate due to the continuously changing angle of the slope and the non-constant acceleration along the path.
To analyze the situation, we apply the principle of energy conservation, which states that the total mechanical energy in a closed system remains constant if only conservative forces are acting. The energy conservation equation can be expressed as:
$$K_{\text{initial}} + U_{\text{initial}} + W_{\text{non-conservative}} = K_{\text{final}} + U_{\text{final}}$$
In this case, the block starts at the ground level, where its initial gravitational potential energy \(U_{\text{initial}}\) is zero. The work done by non-conservative forces \(W_{\text{non-conservative}}\) is also zero, as there are no applied forces or friction acting on the block. At the top of the hill, the final kinetic energy \(K_{\text{final}}\) is zero because we are interested in the minimum speed required to just reach the top without any additional kinetic energy.
Thus, the equation simplifies to:
$$\frac{1}{2} m v_a^2 = m g y_b$$
Here, \(v_a\) is the initial speed, \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)), and \(y_b\) is the height of the hill (20 meters). By canceling the mass \(m\) from both sides, we can rearrange the equation to solve for the minimum speed:
$$v_a = \sqrt{2 g y_b}$$
Substituting the known values, we find:
$$v_a = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 20 \, \text{m}}$$
Calculating this gives:
$$v_a \approx 19.8 \, \text{m/s}$$
This result indicates that the minimum speed required for the block to reach the top of the hill is approximately 19.8 meters per second. Understanding this concept is crucial for solving similar problems involving energy conservation in curved paths.
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