Question

1. use the image to answer the question. a 1.5 m long frictionless pendulum of mass 1.6 kg is released from point a at an angle θ of 10 degrees. what is the speed of the pendulum at point d, half the height up to point e? (1 point) 0.47 m/s 0.67 m/s 0.36 m/s 0.18 m/s

Explanation:

Step1: Calculate the height difference

The height of point \(A\) is \(h = L(1 - \cos\theta)\), where \(L = 1.5\space m\) and \(\theta=10^{\circ}\).

$$h = 1.5\times(1-\cos10^{\circ})$$

$$h\approx1.5\times(1 - 0.9848)=1.5\times0.0152 = 0.0228\space m$$

The height of point \(D\) is \(\frac{h}{2}\).

Step2: Apply conservation of mechanical energy

By the conservation of mechanical energy \(mgh=\frac{1}{2}mv^{2}+mg\frac{h}{2}\) (taking the lowest - point as the zero - potential level).
Cancel out \(m\) (mass) from both sides of the equation: \(gh=\frac{1}{2}v^{2}+g\frac{h}{2}\)
Rearrange the equation to solve for \(v\): \(\frac{1}{2}v^{2}=g\frac{h}{2}\)

$$v = \sqrt{gh}$$

Substitute \(g = 9.8\space m/s^{2}\) and \(h = 0.0228\space m\)

$$v=\sqrt{9.8\times0.0228}$$

$$v=\sqrt{0.22344}\approx0.47\space m/s$$

Answer:

0.47 m/s