A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

Question

Aleksandr

3 months ago

10

A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

e the ground? What is its speed at that point?

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-02-28T18:18:56+03:00

Solution:

The kinetic energy of the clam at an elevation of 5.0 m is 5.19 J and the velocity of the clam at that height is 9.71 m/s.

Explanation:

Throughout its motion, mechanical energy remains constant. We understand that mechanical energy is the summation of potential energy and kinetic energy. Potential energy = $m \times g \times h$ , Kinetic energy = $\frac{1}{2} \times m \times v^{2}$ and Mechanical energy = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Initial kinetic energy is zero. At a height of 9.8 m, the mechanical energy of the clam with a mass of 0.11 kg and g=9.81 $\frac{m}{s^{2}}$ is calculated as follows: 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of the clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ = $5.39+\frac{1}{2} \times m \times v^{2}$ . Given that mechanical energy is conserved, we can state that the mechanical energy of the clam at a height of 9.8 m is equal to that at 5.0 m. The representation is as follows:

10.58 = $5.39+\frac{1}{2} \times m \times v^{2}$ 10.58 – 5.39 = $\frac{1}{2} \times m \times v^{2}$ 5.19 = $\frac{1}{2} \times m \times v^{2}$ the clam's kinetic energy measures 5.19 J.

Lastly, the speed of the clam at 5.0 m is computed; thus, 5.19 = $\frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2}$ 94.36 = $v^{2} \sqrt{94.36}=v \quad v$ = 9.71 m/s. The clam's speed is determined to be 9.71 m/s.

A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

1/7 kg