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You are testing a new amusement park roller coaster with an empty car of mass 120 kg. One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop (point \(A\)) the car has speed 25.0 m/s, and at the top of the loop (point \(B\)) it has speed 8.0 m/s. As the car rolls from point \(A\) to point \(B\), how much work is done by friction?

Short Answer

Expert verified
The work done by friction is -5460 J.

Step by step solution

01

Identify Known Values

List the known values from the problem: - Mass of the car, \( m = 120 \, \text{kg} \)- Radius of the loop, \( r = 12.0 \, \text{m} \)- Speed at the bottom of the loop (point \( A \)), \( v_A = 25.0 \, \text{m/s} \)- Speed at the top of the loop (point \( B \)), \( v_B = 8.0 \, \text{m/s} \)- Height difference between point \( A \) and \( B \), \( h = 2r = 24.0 \, \text{m} \).
02

Calculate Initial and Final Kinetic Energy

Use the formula for kinetic energy: \( KE = \frac{1}{2}mv^2 \).- Initial kinetic energy at point \( A \): \[ KE_A = \frac{1}{2} \times 120 \, \text{kg} \times (25.0 \, \text{m/s})^2 = 37500 \, \text{J} \]- Final kinetic energy at point \( B \): \[ KE_B = \frac{1}{2} \times 120 \, \text{kg} \times (8.0 \, \text{m/s})^2 = 3840 \, \text{J} \]
03

Calculate Potential Energy at Point B

Use the formula for gravitational potential energy: \( PE = mgh \), where \( g \approx 9.81 \, \text{m/s}^2 \).- Potential energy at point \( B \): \[ PE_B = 120 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 24.0 \, \text{m} = 28202.4 \, \text{J} \]
04

Apply the Work-Energy Principle

According to the work-energy principle, the work done by friction \( W_f \) is the change in mechanical energy (kinetic + potential) from point \( A \) to point \( B \):\[W_f = (KE_B + PE_B) - KE_A\]Substitute the values:\[W_f = (3840 \, \text{J} + 28202.4 \, \text{J}) - 37500 \, \text{J}\]\[W_f = 32042.4 \, \text{J} - 37500 \, \text{J} = -5460 \, \text{J}\]
05

Interpret the Result

The work done by friction is \(-5460 \, \text{J}\), indicating that friction has removed this amount of energy from the car's mechanical energy during the motion from point \( A \) to point \( B \). The negative sign shows that friction does work against the motion.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of a body due to its motion. This form of energy is directly tied to the velocity and mass of the moving object. In our roller coaster example, kinetic energy is a crucial aspect since the car's speed changes as it moves through the loop.

The formula for kinetic energy is given by:
  • \( KE = \frac{1}{2}mv^2 \)
Where \( m \) is the mass of the car and \( v \) is its velocity.

At the bottom of the loop (point \( A \)), the car has a higher speed, resulting in greater kinetic energy. This is calculated as 37,500 J based on the car's mass (120 kg) and velocity (25.0 m/s). As the car ascends and reaches the top of the loop (point \( B \)), its speed decreases to 8.0 m/s, reducing its kinetic energy to 3,840 J. The change in kinetic energy is part of the energy analysis to understand how energy transitions through different states in the system.
Potential Energy
Potential energy is the stored energy of an object because of its position or state. For an object elevated above a reference point, this form of energy is known as gravitational potential energy.

The calculation for gravitational potential energy is:
  • \( PE = mgh \)
Where \( m \) is the mass, \( g \) is the acceleration due to gravity (approx. 9.81 \( \text{m/s}^2 \)), and \( h \) is the height above ground.

In the roller coaster track, as the car moves from the bottom to the top of the loop, it gains height equal to twice the radius of the loop (24.0 m). The potential energy at the top becomes 28,202.4 J. This gained potential energy comes from the work done against gravity, effectively transforming some of the car's kinetic energy into potential energy as it climbs up the loop.
Friction
Friction is a force that opposes the relative motion of two surfaces in contact. It converts kinetic energy into other forms, primarily heat, and reduces the mechanical energy of systems in motion.

In our scenario, as the roller coaster car moves from point \( A \) to point \( B \), friction performs work on the car. This work results in a loss of 5,460 J from the overall mechanical energy, as evidenced by the drop in kinetic energy and the work-energy principle calculations.

The frictional force reduces the car's speed and kinetic energy while dissipating energy as heat. This effect can be observed in many real-world systems, where understanding the impact of friction is crucial to manage energy efficiency and performance.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system. It highlights the interplay between active motion and stored energy due to position.

For the roller coaster car, at point \( A \), the mechanical energy is purely kinetic since the car is at a lower height. As it progresses to point \( B \), some kinetic energy transitions into potential energy, comprising the total mechanical energy at that point.

The work-energy principle helps us track these changes by defining the work done, such as by friction, impacting the mechanical energy of the system. As the car loses 5,460 J of energy due to friction, the total mechanical energy is reduced from its initial state. This concept is essential for analyzing energy conservation and loss in real-world mechanical systems.

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Most popular questions from this chapter

A 0.60-kg book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.8 N. (a) How much work is done on the book by friction during a displacement of 3.0 m to the left? (b) The book now slides 3.0 m to the right, returning to its starting point. During this second 3.0-m displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

A force parallel to the \(x\)-axis acts on a particle moving along the x-axis. This force produces potential energy \(U(x)\) given by \(U(x) = \alpha x^4\), where \(\alpha =\) 0.630 J/m\(^4\). What is the force (magnitude and direction) when the particle is at \(x = -0.800\) m?

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{F}\) \(= -a x^2 \hat{\imath}\), where \(\alpha = 12 \mathrm{N/m}^2\). (a) How much work does \(\overrightarrow{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\overrightarrow{F}\) conservative? Explain. If \(\overrightarrow{F}\) is conservative, what is the potential-energy function for it? Let \(U =\) 0 when \(x =\) 0.

A force of 520 N keeps a certain spring stretched a distance of 0.200 m. (a) What is the potential energy of the spring when it is stretched 0.200 m? (b) What is its potential energy when it is compressed 5.00 cm?

A truck with mass \(m\) has a brake failure while going down an icy mountain road of constant downward slope angle \(\alpha\) (\(\textbf{Fig. P7.58}\)). Initially the truck is moving downhill at speed \(v_0\). After careening downhill a distance \(L\) with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle \(\beta\). The truck ramp has a soft sand surface for which the coefficient of rolling friction is \(\mu_r\). What is the distance that the truck moves up the ramp before coming to a halt? Solve by energy methods.

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