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4. what is the potential energy of an 839 g coconut that is at the top …

Question

  1. what is the potential energy of an 839 g coconut that is at the top of a 25 m tall tree?
  2. a 1.4 kg ball has 55 j of potential energy. how high from the ground is the ball?
  3. one of the tallest radio towers is in fargo, north dakota. the tower is 629.9 m tall. if a bird lands on top of the tower, so that the gravitational potential energy associated with the bird is 2033.76 j, what is its mass, in kilograms?

Explanation:

Step1: Convert mass unit

The formula for gravitational potential energy is \(U = mgh\), where \(g = 9.8\ m/s^{2}\). First, convert the mass of the coconut from grams to kilograms. \(m=839\ g=0.839\ kg\).

Step2: Calculate potential energy

Substitute \(m = 0.839\ kg\), \(g = 9.8\ m/s^{2}\), and \(h = 25\ m\) into the formula \(U = mgh\).
\(U=0.839\times9.8\times25\)
\(U = 0.839\times245\)
\(U=205.555\ J\)

Step3: Solve for height (for the ball problem)

Given \(U = 55\ J\), \(m = 1.4\ kg\), \(g = 9.8\ m/s^{2}\), from \(U = mgh\), we can solve for \(h\). \(h=\frac{U}{mg}\)
\(h=\frac{55}{1.4\times9.8}\)
\(h=\frac{55}{13.72}\approx4.01\ m\)

Step4: Solve for mass (for the bird problem)

Given \(U = 2033.76\ J\), \(h = 629.9\ m\), \(g = 9.8\ m/s^{2}\), from \(U = mgh\), we can solve for \(m\). \(m=\frac{U}{gh}\)
\(m=\frac{2033.76}{9.8\times629.9}\)
\(m=\frac{2033.76}{6173.02}\approx0.33\ kg\)

Answer:

  1. \(205.555\ J\)
  2. \(4.01\ m\)
  3. \(0.33\ kg\)