Kathy is changing the tire of her car on a steep hill 20m high. she trips and drops the 10kg spare tire which rolls down the hill. what is the speed of the tire at the top of the next hill if the height of the hill is 5m high?

For my answer to the question above, Let's assume that there is no friction.

The total energy E (kinetic + potential) of the tire is: 

E = mv^2 / 2 + mgh 

Since we're not given the tire's mass distribution or the hill's coefficient of friction or anything, assume we're to disregard rotational inertia, the energy dissipated as heat, etc. 

Given that: 

m = 10.0 kg 
v0 = 2.0 m/s 
h0 = 20.0 m 

We're to find v when h = 5.0 m 

Since the overall energy is conserved, 

m(v0)^2 / 2 + mg(h0) = mv^2 / 2 + mgh 

=> 

v = sqrt((v0)^2 + 2g(h0 - h) 

= sqrt( 4.0 m^2 / s^2 + (2)(9.8 m / s^2) (15.0 m) ) 

= sqrt( 298 m^2 / s^2 ) 

= 17.26 m/s

B) as the temperature increases, the kinetic energy of the particles increases.|-/