During a rockslide, a 740 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.18.

(a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill?(d) What is its speed then?

(a) Well, since the gravitational potential energy U is set to zero at the bottom of the hill, just before the slide it can be calculated using the formula U = mgh, where m is the mass of the rock, g is the acceleration due to gravity, and h is the height of the hill. Plug in the values and do the math, and you'll have your answer. But hey, don't be too hard on the rock, it's just hanging around before its big adventure!

(b) Ah, thermal energy, the land where all energy goes to retire! During the slide, some of the energy gets transferred to thermal energy due to the friction between the rock and the hill surface. But don't worry, it's essentially the rock's way of giving the hill a warm hug. To find out how much energy is transferred, you can use the formula Energy Transferred = Work = Force of Friction * Distance. It's like a little frictional picnic, but hotter!

(c) As the rock reaches the bottom of the hill, it gains kinetic energy, ready to rock and roll! To calculate this energy, you can use the formula Kinetic Energy = (1/2)mv^2, where m is the mass of the rock and v is its velocity. It's the rock's way of saying, "I've got energy to spare!"

(d) The speed of the rock at the bottom of the hill can be found using the equation v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the hill. Once you find the speed, you'll realize the rock is really on a roll!

To solve this problem, we can break it down into several steps:

Step 1: Calculate the gravitational potential energy just before the slide (U).
Step 2: Determine the work done by friction (Wfriction).
Step 3: Calculate the energy transferred to thermal energy during the slide (Q).
Step 4: Calculate the remaining energy as kinetic energy (K).
Step 5: Determine the speed of the rock at the bottom of the hill (v).

Let's go through each step one by one.

Step 1: Calculate the gravitational potential energy just before the slide (U).

The formula for gravitational potential energy is U = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Given:
Mass of the rock (m) = 740 kg
Height of the hill (h) = 300 m
Acceleration due to gravity (g) = 9.8 m/s²

Using the formula U = mgh, we can calculate the potential energy just before the slide:

U = (740 kg) * (9.8 m/s²) * (300 m)
U = 2,150,400 J

Therefore, the gravitational potential energy just before the slide is 2,150,400 J.

Step 2: Determine the work done by friction (Wfriction).

The formula for work done by friction is Wfriction = µ * m * g * d, where µ is the coefficient of kinetic friction, m is the mass, g is the acceleration due to gravity, and d is the distance.

Given:
Coefficient of kinetic friction (µ) = 0.18
Mass of the rock (m) = 740 kg
Distance of the hill (d) = 500 m
Acceleration due to gravity (g) = 9.8 m/s²

Using the formula Wfriction = µ * m * g * d, we can calculate the work done by friction:

Wfriction = (0.18) * (740 kg) * (9.8 m/s²) * (500 m)
Wfriction = 6,747,600 J

Therefore, the work done by friction is 6,747,600 J.

Step 3: Calculate the energy transferred to thermal energy during the slide (Q).

The energy transferred to thermal energy (Q) is equal to the work done by friction, since friction converts mechanical energy into heat.

Therefore, Q = Wfriction = 6,747,600 J.

Step 4: Calculate the remaining energy as kinetic energy (K).

The remaining energy (K) is equal to the initial potential energy (U) minus the energy transferred to thermal energy (Q).

Therefore, K = U - Q = 2,150,400 J - 6,747,600 J = -4,597,200 J

The negative value for K indicates that some of the initial energy was transferred to thermal energy.

Step 5: Determine the speed of the rock at the bottom of the hill (v).

The formula for kinetic energy is K = 0.5 * m * v², where m is the mass and v is the velocity.

Given:
Mass of the rock (m) = 740 kg
Kinetic energy (K) = -4,597,200 J

Rearranging the formula, we get v² = (2K) / m.

v² = (2 * -4,597,200 J) / (740 kg)
v² = -9,194,400 J / (740 kg)
v² = -12,416.2162 m²/s²

Taking the square root, we find the speed of the rock at the bottom of the hill (v):

v = √(-12,416.2162 m²/s²)
v = 111.33 m/s (rounded to 2 decimal places)

Therefore, the speed of the rock at the bottom of the hill is 111.33 m/s (rounded to 2 decimal places).

To summarize:
(a) The value of U just before the slide is 2,150,400 J.
(b) The energy transferred to thermal energy during the slide is 6,747,600 J.
(c) The kinetic energy of the rock as it reaches the bottom of the hill is -4,597,200 J.
(d) The speed of the rock at the bottom of the hill is 111.33 m/s (rounded to 2 decimal places).

To answer these questions, we need to apply the principles of work-energy theorem and conservation of energy. Let's break it down into steps:

(a) To find the value of U just before the slide, we need to find the change in gravitational potential energy (ΔU). The gravitational potential energy of an object of mass m and height h is given by U = mgh, where g is the acceleration due to gravity.

In this case, the rock is at rest at the top of the hill, so its initial gravitational potential energy is zero at the top. Therefore, the value of U just before the slide is also zero.

(b) To find the amount of energy transferred to thermal energy during the slide, we need to find the work done by frictional force.

The work done by the frictional force can be calculated using the formula: Wfriction = -μk * N * d, where μk is the coefficient of kinetic friction, N is the normal force, and d is the distance. The negative sign indicates that the work done by friction is in the opposite direction of motion.

The normal force can be calculated using N = mg, where m is the mass of the rock and g is the acceleration due to gravity.

Substituting the given values, we have: N = (740 kg) * (9.8 m/s^2) = 7252 N.

The distance is given as 500 m.

Wfriction = - (0.18) * (7252 N) * (500 m)

(c) To find the kinetic energy of the rock as it reaches the bottom of the hill, we can use the conservation of mechanical energy. The total mechanical energy of the system remains constant if no external work is done. Therefore, the sum of the initial kinetic energy and the initial potential energy should be equal to the sum of the final kinetic energy and the final potential energy.

The initial kinetic energy is zero since the rock starts from rest. The initial potential energy is given by U = mgh, where h is the height.

Substituting the given values: U = (740 kg) * (9.8 m/s^2) * (300 m)

The final potential energy is zero at the bottom of the hill since that's where the reference point for potential energy is chosen.

Therefore, the kinetic energy at the bottom of the hill is equal to the initial potential energy.

(d) To find the speed of the rock at the bottom of the hill, we can use the kinetic energy.

The kinetic energy is given by KE = 1/2 * mv^2, where m is the mass of the rock and v is the speed.

Setting the initial potential energy equal to the kinetic energy, we have:

(740 kg) * (9.8 m/s^2) * (300 m) = 1/2 * (740 kg) * v^2

Simplifying the equation and solving for v will give us the speed of the rock at the bottom of the hill.