Question

1. use the image to answer the question. a frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point a, at an angle θ of 60°. as it swings downward, it passes through point b, which is 30 degrees from equilibrium. what is the kinetic energy of the pendulum at point b? (1 point) 4.1 j 1.1 j 3.9 j 3.0 j

Explanation:

Step1: Calculate height at point A

The height \( h_A\) of the pendulum at point A (angle \(\theta = 60^{\circ}\)) is given by \(h_A = L(1 - \cos\theta)\), where \(L = 2.1\ m\).

$$ h_A=2.1\times(1 - \cos60^{\circ})=2.1\times(1 - 0.5)=1.05\ m $$

Step2: Calculate height at point B

The height \( h_B\) of the pendulum at point B (angle \(\alpha = 30^{\circ}\)) is given by \(h_B = L(1 - \cos\alpha)\)

$$ h_B=2.1\times(1 - \cos30^{\circ})=2.1\times(1-\frac{\sqrt{3}}{2})\approx2.1\times(1 - 0.866)=0.2874\ m $$

Step3: Apply conservation of mechanical energy

By the law of conservation of mechanical energy \(E = K + U\) (where \(E\) is total mechanical energy, \(K\) is kinetic energy and \(U\) is potential energy). At point A, \(K_A = 0\) (starts from rest), so \(E = U_A=mgh_A\). At point B, \(E = K_B+U_B=mgh_B + K_B\). Then \(K_B=mg(h_A - h_B)\)
Given \(m = 0.4\ kg\), \(g = 9.8\ m/s^{2}\), \(h_A - h_B=1.05 - 0.2874 = 0.7626\ m\)

$$ K_B=0.4\times9.8\times0.7626\approx3.0\ J $$

Answer:

3.0 J