Question

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

Answer 1

Answer:

The kinetic energy of the clam at a height of 5.0 m is 5.19 J and the speed of the clam at that height is 9.71 m/s.

Explanation:

Mechanical energy is constant throughout the travel, we know that mechanical energy is calculated by adding 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 = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Kinetic energy is zero at initial point. Now mechanical energy of clam with m=0.11kg,g=9.81$\frac{m}{s^{2}}$,h=9.8 m is = 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ =$5.39+\frac{1}{2} \times m \times v^{2}$. We know that mechanical energy is constant hence, mechanical energy of clam at height 9.8 m is equal to mechanical energy at height 5.0 m. This is represented as following

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 = $\frac{1}{2} \times m \times v^{2}$ kinetic energy of the clam is 5.19 J.

Now speed of the clam at height 5.0 m is 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 speed of the clam is 9.71 m/s.