Question

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?

Answer 1

The speed of the tire at the top of the 5m hill, calculated using conservation of energy principles and ignoring any work done by friction, is approximately 17.15 m/s.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy principle, which states that if no external work is done on the system (like work by friction), the total mechanical energy remains constant. This means that the potential energy lost by the tire as it rolls down from the higher hill will be converted into kinetic energy.

The potential energy at the top of the 20m hill is given by PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is the height of the hill. At the 20m hill, PE = 10kg × 9.8 m/s2 × 20m. When the tire reaches the top of the next hill, its potential energy will be PE = 10kg × 9.8 m/s2 × 5m.

We can then equate the initial potential energy minus the final potential energy to the kinetic energy at the top of the 5m hill: KE = ½ mv2, and solve for the speed v.

Conservation of energy: mgh1 - mgh2 = ½ mv2

Calculation:

PE at 20m: (10 × 9.8 × 20) J = 1960 J

PE at 5m: (10 × 9.8 × 5) J = 490 J

Kinetic energy at 5m hill: 1960 J - 490 J = 1470 J

1470 J = ½ × 10kg × v2

v2 = (1470 J × 2) / 10kg

v2 = 294 m2/s2

v = √294 m2/s2

v ≈ 17.15 m/s

Therefore, the speed of the tire at the top of the 5m hill is approximately 17.15 m/s.

Answer 2

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