Question

A 17-kg rock is on the edge of a 103-m cliff.

(a) What potential energy does the rock possess relative to the base of the cliff? J
(b) The rock falls from the cliff. What is its kinetic energy just before it strikes the ground? J
(c) What speed does the rock have as it strikes the ground?

Answer 1

the guy above me is correct

Step-by-step explanation:

Answer 2

a)
`U=17,159.8J`

b)
`K=17,159.8J`

c)
`v=44.93m/s`

Step-by-step explanation:

The information we have is

mass:
`m=17kg`

height:
`h=103m`

and we also know the acceleration of gravity is:
`g=9.8m/s^2`

a) the potential energy is:

`U=mgh\\U=(17kg)(9.8m/s^2)(103m)=17,159.8J`

b) let's call the potential energy at the top of the cliff
`U` and at the bottom
`U_(2)`, and call the kinetic energy at the top of the cliff
`K` and at the bottom of the cliff
`K_(2)`. The law of conservation of energy tells us that:

`U+K=U_(2)+K_(2)`

and because at the top of the cliff the rock does not move the kinetic energy there is zero.
`K=0`. In addition, when reaching the bottom of the cliff the height is
`h=0`, so that the potential energy at the bottom is zero:
`U_(2)=0`

so:

`U+0=0+K_(2)`

`U=K_(2)`

The kinetic energy when the stone touches the ground is the same as the potential energy at the top of the cliff:

`K=17,159.8J`

c) to find the speed we will use the formula for kinetic energy

`K=(1)/(2) mv^2`

clearing for the speed:

`v=\sqrt{(2K)/(m) }`

and we already know the kinetic energy as the rock strikes the ground:
`K=17,159.8J`

so the speed:

`v=\sqrt{(2(17,159.8J))/((17kg)) }`

`v=\sqrt{(34,319.6J)/(17kg) }`

`v=√(2,018.8J/kg)`

`v=44.93m/s`