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During a rockslide, a \(520 \mathrm{~kg}\) rock slides from rest down a hillside that is \(500 \mathrm{~m}\) long and \(300 \mathrm{~m}\) high. The coefficient of kinetic friction between the rock and the hill surface is \(0.25 .\) (a) If the gravitational potential energy \(U\) of the rock-Earth system is 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?

Short Answer

Expert verified
(a) 1,528,800 J; (b) 508,000 J; (c) 1,020,800 J; (d) 62.57 m/s.

Step by step solution

01

Calculate Initial Gravitational Potential Energy

To calculate the initial gravitational potential energy, use the formula \[ U = mgh \] where \( m = 520 \; \text{kg} \), \( g = 9.8 \; \text{m/s}^2 \), and \( h = 300 \; \text{m} \). Substitute the values to get:\[ U = 520 \times 9.8 \times 300 \].Multiplying these values gives \( U = 1,528,800 \; \text{J} \).
02

Calculate Energy Transferred to Thermal Energy

The work done by friction which is equal to the energy transferred to thermal energy \( E_{th} \) can be calculated by\[ E_{th} = f_k \cdot d \],where \( f_k = \mu_k \cdot N \). \( N = mg \cos(\theta) \) is the normal force.First, we find \( \sin(\theta) = 300 / 500 = 0.6 \) and hence \( \cos(\theta) = 0.8 \).So, \( N = 520 \times 9.8 \times 0.8 = 4064 \; \text{N} \).Then, \( f_k = 0.25 \times 4064 = 1016 \; \text{N} \).Thus, \( E_{th} = 1016 \times 500 = 508,000 \; \text{J} \).
03

Calculate Kinetic Energy at Bottom of Hill

The total mechanical energy at the beginning is converted into kinetic energy and thermal energy. So we use the conservation of energy:\[ K + E_{th} = U \].Rearranging the formula gives:\[ K = U - E_{th} = 1,528,800 - 508,000 \].Thus, kinetic energy \( K = 1,020,800 \; \text{J} \).
04

Calculate Speed at Bottom of Hill

Use the formula for kinetic energy:\[ K = \frac{1}{2} mv^2 \].Rearrange to find \( v \):\[ v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1,020,800}{520}} \].Solving this gives \( v \approx 62.57 \; \text{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its height above the ground, which is influenced by gravity. It is calculated using the formula \( U = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration (approximately \(9.8\, \text{m/s}^2\) on Earth), and \( h \) is the height. For example, if a \(520\, \text{kg}\) rock is sitting on a hill \(300\, \text{m}\) high, its gravitational potential energy is multiplied by these values and results in \(1,528,800\, \text{J}\). This energy represents the potential for the rock to do work as it moves down the hill, such as converting this potential energy into kinetic energy.
Kinetic Friction
Kinetic friction occurs when two surfaces slide past each other, opposing the direction of motion. It is characterized by the coefficient of kinetic friction, \( \mu_k \), which varies depending on the materials in contact. In our scenario, the coefficient of kinetic friction between the rock and the hill is \(0.25\). To calculate the work done by friction, we determine the normal force, \( N \), by multiplying the gravitational force by the cosine of the angle of the slope. The frictional force, \( f_k \), is then \( \mu_k \times N \). Here, \( N = 4064\, \text{N} \) gives a frictional force of \(1016\, \text{N}\), resulting in \(508,000\, \text{J}\) of energy transformed into thermal energy as the rock slides down the hill.
Thermal Energy
As the rock slides down the hill, kinetic friction converts some of its mechanical energy into thermal energy. This transformation is due to the frictional force acting against the movement. Thermal energy in this context is basically the heat produced by the friction between the rock and the hill's surface. In the exercise, the friction causes significant energy loss, transforming \(508,000\, \text{J}\) of the rock's initial mechanical energy into thermal energy. This energy is irrecoverable and disperses into the surroundings as heat, demonstrating an essential aspect of energy dissipation in mechanical systems.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, calculated using \( K = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. For the rock in motion, its kinetic energy increases as gravitational potential energy and some converted thermal energy combine. By the end of its downhill journey, the rock's kinetic energy is \(1,020,800\, \text{J}\), achieved using the initial potential energy minus the thermal energy dissipated during the slide. This high kinetic energy allows us to also determine the final speed of the rock, which is approximately \(62.57\, \text{m/s}\). Understanding kinetic energy is critical in grasping how efficiently an object's potential energy is transformed as it moves.
Mechanical Energy Conservation
Mechanical energy conservation is a fundamental principle stating that the total mechanical energy (the sum of potential and kinetic energy) in a closed system remains constant, provided no external work is done. However, with non-conservative forces like friction, energy can change forms. In the exercise, the rock's initial gravitational potential energy is partially transformed into kinetic and thermal energy due to friction. The loss in mechanical energy appears as heat energy, indicating that while the form of energy changes, the total energy within the system (including thermal energy) is constant. This demonstrates how energy conservation is applied in real-world scenarios, emphasizing energy transformations rather than loss.

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Most popular questions from this chapter

The only force acting on a particle is conservative force \(\vec{F}\). If the particle is at point \(A,\) the potential energy of the system associated with \(\vec{F}\) and the particle is \(40 \mathrm{~J}\). If the particle moves from point \(A\) to point \(B,\) the work done on the particle by \(\vec{F}\) is \(+25 \mathrm{~J}\). What is the potential energy of the system with the particle at \(B ?\)

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

A skier weighing \(600 \mathrm{~N}\) goes over a frictionless circular hill of radius \(R=20 \mathrm{~m}\) (Fig. 8 -62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is \(8.0 \mathrm{~m} / \mathrm{s}\) at point \(B,\) at angle \(\theta=20^{\circ} .\) (a) What is her speed at the hilltop (point \(A\) ) if she coasts without using her poles? (b) What minimum speed can she have at \(B\) and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs \(700 \mathrm{~N}\) instead of \(600 \mathrm{~N} ?\)

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is 0.10 . (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s},\) what is her speed at the bottom?

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. \(8-37\) ) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

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