Question

A crow drops .11 -kg clam onto a rocky beach from a height of 9.8 m. What is the kinetic energy of the clam when it is 5.0 m above the ground? What is it's speed at that point?

Answer 1

Answer:

Kinetic Energy=5.17J

Speed=9.7m/sec

Explanation:

Given the mass of the clam is m=0.11kg

The height from which the rock was thrown is h=9.8m

Given the rock is at 5m above the ground.

The distance travelled by the rock is (s)=9.8-5=4.8m

The initial velocity is u=0m/sec

let the final velocity be v m/sec

We know that the acceleration of the particle is $a=9.8m/sec^{2}$

We know that $v^{2} -u^{2} =2as$

$v^{2} -0^{2} =2\times 9.8\times 4.8$

$v^{2} =94.08$

v=9.7m/sec

Now the kinetic energy=$\frac{1}{2}mv^{2}$

KE=$\frac{1}{2}\times 0.11\times 9.7^{2}$=5.17J