A sled and rider with a combined weight of 59 kg are at rest on the top of a hill 24 m high.

(a) What is their total energy at the top of the hill?
(b) Assuming there is no friction, what would the total energy be on sliding halfway down the hill?

a. PE max = Mg*h = 59*9.8*24 = 13,877 J.

b. KE + PE = PE max = 13,877 J.

3550J

(a) What is their total energy at the top of the hill?

Well, let's crunch the numbers! The total energy at the top of the hill can be calculated using the equation:

Potential energy = mgh

where m is the combined weight of the sled and rider (59 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the hill (24 m).

Potential energy = 59 kg * 9.8 m/s^2 * 24 m = 13,150 J

So, their total energy at the top of the hill is a whopping 13,150 Joules!

(b) Assuming there is no friction, what would the total energy be on sliding halfway down the hill?

Well, halfway down the hill, the potential energy will decrease, but the kinetic energy will increase. Since there is no friction involved, we can assume that all the loss in potential energy is converted into kinetic energy.

Therefore, at the halfway point, the potential energy becomes zero, and the total energy is solely kinetic energy. So, the total energy would be equal to the kinetic energy.

We can calculate the kinetic energy using the equation:

Kinetic energy = (1/2)mv^2

where m is still the combined weight of the sled and rider (59 kg), and v is the velocity at the halfway point. We can find the velocity using the conservation of mechanical energy, which states that the total energy remains constant.

Total energy = Potential energy + Kinetic energy

Since the potential energy is zero halfway down the hill, the total energy equals the kinetic energy.

13,150 J + 0 = (1/2) * 59 kg * v^2

0.5 * 59 kg * v^2 = 13,150 J

Now, the v^2 isolates on one side:

v^2 = (13,150 J) / (0.5 * 59 kg)

v^2 ≈ 445.76 m^2/s^2

And taking the square root of both sides:

v ≈ √(445.76 m^2/s^2)

v ≈ 21.1 m/s

So, on sliding halfway down the hill, the total energy, which is solely kinetic energy, would be approximately 21.1 Joules.

To answer these questions, we need to understand the concept of potential energy and kinetic energy.

(a) The total energy at the top of the hill is the sum of the potential energy and the rider's kinetic energy. The potential energy is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the hill.

Substituting the given values: m = 59 kg and h = 24 m, we have:

PE = 59 kg * 9.8 m/s^2 * 24 m
= 13846.4 J

The potential energy at the top of the hill is 13846.4 Joules.

(b) Assuming no friction, the potential energy at the halfway point is converted entirely to kinetic energy. The kinetic energy is given by the formula KE = 1/2 * m * v^2, where m is the mass and v is the velocity.

To find the velocity halfway down the hill, we can use the principle of conservation of energy. The potential energy at the top of the hill is equal to the sum of the potential energy and kinetic energy halfway down the hill. Therefore:

PE (at the top of the hill) = PE (halfway down) + KE (halfway down)

Substituting the values: PE = 13846.4 J and h = 24 m, we have:

13846.4 J = m * g * (h/2) + 1/2 * m * v^2

Since the sled and rider have a combined weight of 59 kg, we substitute m = 59 kg:

13846.4 J = 59 kg * 9.8 m/s^2 * (24 m/2) + 1/2 * 59 kg * v^2

Simplifying:

13846.4 J = 7058.8 J + 29.05 * v^2

Rearranging the equation:

6796.2 J = 29.05 * v^2

Dividing by 29.05:

v^2 = 233.957

Taking the square root of both sides:

v ≈ 15.29 m/s

The velocity halfway down the hill is approximately 15.29 m/s.

The total energy halfway down the hill (assuming no friction) is the sum of the potential energy and kinetic energy:

Total Energy = PE + KE
= 59 kg * 9.8 m/s^2 * (24 m/2) + 1/2 * 59 kg * (15.29 m/s)^2

Simplifying:

Total Energy ≈ 7058.8 J + 1/2 * 59 kg * (15.29 m/s)^2

Total Energy ≈ 7559.2 J

Therefore, the total energy halfway down the hill (assuming no friction) is approximately 7559.2 Joules.