Question

use the image to answer the question.
a frictionless pendulum has a mass of 1.7 kg and a length of 2.2 m. if the pendulum is released from point a and attains a speed of 2.9 m/s at point c, then how high will the pendulum rise at point d, halfway to its maximum height?
(1 point)
3.6 m
7.1 m
0.43 m
0.21 m

Explanation:

Step1: Apply conservation of mechanical energy

According to the law of conservation of mechanical energy \(E = E_{k}+E_{p}\), where \(E_{k}=\frac{1}{2}mv^{2}\) and \(E_{p}=mgh\). At point \(A\), \(E = mgh\) (taking the lowest - point \(C\) as the zero - potential level). At point \(C\), \(E=\frac{1}{2}mv_{C}^{2}\). So \(mgh=\frac{1}{2}mv_{C}^{2}\), and \(h=\frac{v_{C}^{2}}{2g}\). Given \(v_{C} = 2.9m/s\) and \(g = 9.8m/s^{2}\), then \(h=\frac{(2.9)^{2}}{2\times9.8}=\frac{8.41}{19.6}\approx0.43m\).

Step2: Find the height at point \(D\)

Since point \(D\) is halfway to the maximum height. The maximum height \(h_{max}\) (from the zero - potential level) is found from the energy at \(C\) (using conservation of energy \(mgh_{max}=\frac{1}{2}mv_{C}^{2}\)). The height at point \(D\), \(h_{D}=\frac{h_{max}}{2}\). Since \(h_{max}=\frac{v_{C}^{2}}{2g}\), then \(h_{D}=\frac{v_{C}^{2}}{4g}\). Substitute \(v_{C} = 2.9m/s\) and \(g = 9.8m/s^{2}\) into the formula: \(h_{D}=\frac{(2.9)^{2}}{4\times9.8}=\frac{8.41}{39.2}\approx0.21m\)

Answer:

\(0.21m\) (the fourth option)