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

Answer:

a)$U=17,159.8J$

b)$K=17,159.8J$

c) $v=44.93m/s$

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

$K=\frac{1}{2} mv^2$

clearing for the speed:

$v=\sqrt{\frac{2K}{m} }$

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

so the speed:

$v=\sqrt{\frac{2(17,159.8J)}{(17kg)} }$

$v=\sqrt{\frac{34,319.6J}{17kg} }$

$v=\sqrt{2,018.8J/kg}$

$v=44.93m/s$

Answer 2

Answer:

the guy above me is correct

Explanation: