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A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

Short Answer

Expert verified
Question: Calculate a) the speed of the mass at the bottom of the inclined plane, b) the speed of the mass at the top of the inclined plane, and c) the total work done by friction from the beginning to the end, given the following parameters: - Spring constant (k): 500 N/m - Mass (m): 0.500 kg - Compression in the spring (x): 0.300 m - Inclined plane's length (L): 4.00 m - Inclined plane's angle (θ): 30.0° - Coefficient of Kinetic friction (μ): 0.350 - Horizontal distance to inclined plane (d): 1.50 m

Step by step solution

01

a) Speed of the mass at the bottom of the inclined plane

We will first find the speed of the mass just after leaving the spring and before reaching the inclined plane. To do this, we need to remember that the total mechanical energy (sum of kinetic and potential energies) is conserved when external forces do no work. Potential energy stored in the compressed spring (Us) is: Us = (1/2) * k * x^2 Kinetic energy of the mass after leaving the spring (K1) is: K1 = (1/2) * m * v_1^2 (where v_1 is the speed of the mass at this point) Since no external force is doing work, we have: Us = K1 (1/2) * k * x^2 = (1/2) * m * v_1^2 Now, we can solve for v_1. Next, we should find the loss of kinetic energy due to the frictional force along the horizontal distance d before entering the inclined plane. Work done by friction (Wf) along the horizontal distance is: Wf = -μ * m * g * d (Negative sign because friction opposes the movement) According to the work-energy theorem, the loss of kinetic energy will equal the work done by friction. So, we can write: K1 - K2 = -Wf (where K2 is the kinetic energy just before entering the incline) Now, we can solve for K2 and find the speed of the mass (v_2) at the bottom of the inclined plane: K2 = (1/2) * m * v_2^2
02

b) Speed of the mass at the top of the inclined plane

To find the speed at the top of the inclined plane, we need to consider the gravitational potential energy, kinetic energy, and energy loss due to friction along the inclined plane. First, let's find the height (h) at the top of the inclined plane: h = L * sin(θ) Now we can find the gravitational potential energy at the top of the inclined plane (Ug): Ug = m * g * h Next, let's find the energy loss due to friction along the inclined plane (Wf_incline): Wf_incline = -μ * m * g * L * cos(θ) (Negative sign due to energy loss) According to the work-energy theorem, for the inclined plane portion, we have: K2 + Ug - K3 = -Wf_incline (where K3 is the kinetic energy at the top of the inclined plane) Now, we can solve for K3 and find the speed of the mass (v_3) at the top of the inclined plane: K3 = (1/2) * m * v_3^2
03

c) The total work done by friction

The total work done by friction from the beginning to the end of the mass's motion is the sum of the work done along the horizontal surface and the inclined plane: Wf_total = Wf + Wf_incline By calculating Wf_total, we'll find the total work done by friction through the entire motion of the mass.

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Key Concepts

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

Conservation of Mechanical Energy
When addressing the conservation of mechanical energy, we're talking about one of the fundamental principles of physics which states that in the absence of non-conservative forces, such as friction and air resistance, the total mechanical energy of a system remains constant.

In the example problem, a compressed spring releases its stored potential energy to propel a mass upwards on an incline. Mechanical energy here is the sum of kinetic energy, due to the mass's motion, and potential energy, both elastic potential energy stored in the spring and gravitational potential energy from the height the mass reaches. If we were to ignore friction, we could predict the mass's velocity at any point simply by equating its initial potential energy with its kinetic energy at that point. Consequently, the exercise illustrates how to apply conservation of mechanical energy to determine the mass's speed at the bottom of the inclined plane, considering the friction force acts along the path.
Work-Energy Theorem
The work-energy theorem is key to understanding how forces, like friction, alter a system's mechanical energy. It states that work done on an object is equal to the change in its kinetic energy. In the context of our problem, work is done by the frictional forces as the mass moves across the horizontal surface and up the inclined plane.

To quantify this, work is calculated as the product of the force applied and the distance over which it acts, and in the case of friction, it's always acting opposite to the direction of motion. The friction thus works to reduce the kinetic energy of the mass, showing a direct application of the work-energy theorem. The equation presented, detailing the kinetic energy loss equating to the frictional work, wonderfully exemplifies this theorem in action.
Spring Constant and Potential Energy
Diving into the topic of spring constant and potential energy, the spring constant, denoted as 'k', is a measure of a spring's stiffness. Mathematical expressions involving the spring constant give us insight into the potential energy stored within a compressed or stretched spring, known as elastic potential energy.

The formula \( U_s = \frac{1}{2} k x^2 \) used in the step-by-step solution relates the spring's compression distance 'x' and its constant 'k' to the potential energy it can store. This potential energy is then entirely converted into kinetic energy when the spring is released if no other external forces like friction were present. This perfectly aligns with the conservation of energy concept, where the energy in a closed system shifts from one form to another but is never lost.
Frictional Force and Work
Finally, we explore the concept of frictional force and work. Friction is a resisting force that occurs when two surfaces move across one another, and it always acts in the opposite direction to the motion of the object. In physics problems, this force is often quantified as a product of the coefficient of friction 'μ' and the normal force - for a horizontal surface, this is typically the object's weight.

The work done by friction is calculated as the force due to friction multiplied by the distance over which it acts, and it results in a decrease in the mechanical energy of the system. In the exercise's perspective, this is essential to account for when determining how much the mass's kinetic energy is reduced both across the horizontal surface and as it ascends the incline. This concept encompasses the calculations provided in the step-by-step solution for finding the work done by friction along two different segments of the mass's path.

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Most popular questions from this chapter

A 70.0 -kg skier moving horizontally at \(4.50 \mathrm{~m} / \mathrm{s}\) encounters a \(20.0^{\circ}\) incline. a) How far up the incline will the skier move before she momentarily stops, ignoring friction? b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is \(0.100 ?\)

The energy height, \(H\), of an aircraft of mass \(m\) at altitude \(h\) and with speed \(v\) is defined as its total energy (with the zero of the potential energy taken at ground level) divided by its weight. Thus, the energy height is a quantity with units of length. a) Derive an expression for the energy height, \(H\), in terms of the quantities \(m, h\), and \(v\). b) A Boeing 747 jet with mass \(3.5 \cdot 10^{5} \mathrm{~kg}\) is cruising in level flight at \(250.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10.0 \mathrm{~km} .\) Calculate the value of its energy height. Note: The energy height is the maximum altitude an aircraft can reach by "zooming" (pulling into a vertical climb without changing the engine thrust). This maneuver is not recommended for a 747 , however.

a) If you are at the top of a toboggan run that is \(40.0 \mathrm{~m}\) high, how fast will you be going at the bottom, provided you can ignore friction between the sled and the track? b) Does the steepness of the run affect how fast you will be going at the bottom? c) If you do not ignore the small friction force, does the steepness of the track affect the value of the speed at the bottom?

A block of mass \(5.0 \mathrm{~kg}\) slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a mass of \(4.0 \mathrm{~kg}\) attached to a horizontal spring (with spring constant of \(k=2000.0 \mathrm{~N} / \mathrm{m}\) ), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest? a) \(0.40 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\)

A 1.00 -kg block initially at rest at the top of a 4.00 -m incline with a slope of \(45.0^{\circ}\) begins to slide down the incline. The upper half of the incline is frictionless, while the lower half is rough, with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.300\). a) How fast is the block moving midway along the incline, before entering the rough section? b) How fast is the block moving at the bottom of the incline?

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