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

Kinetic Energy=5.17J

Speed=9.7m/sec

Step-by-step 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* 9.8* 4.8`

`v^(2) =94.08`

v=9.7m/sec

Now the kinetic energy=
`(1)/(2)mv^(2)`

KE=
`(1)/(2)* 0.11* 9.7^(2)`=5.17J