Arock is sliding down a hill. assume that the hill is 20 meters high and the rock has a mass of 1,000 kilograms. if the rock started at the top of the hill with no initial speed, how much is its kinetic energy when it reaches the bottom of the hill? use g=10 m/s2. give the answer in kilojoules.

200 KJ

Explanation:

h = 20 m, m = 1000 kg, g = 10 m/s^2

As the rock is sliding and it starts from rest. it falls from a f=height of 20 m

So, by using the energy conservation law

Potential energy at the top of hill = Kinetic energy at the bottom of hill

So, Kinetic energy at the bottom of hill = Potential energy at the top of hill

K.E = m g h

K. E = 1000 x 10 x 20 = 200,000 J

K.E = 200 KJ

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2