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A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

Short Answer

Expert verified
The final speed of the block is approximately 3.99 m/s.

Step by step solution

01

Identify Known Values

We have a block of ice with mass \( m = 2.00 \, \text{kg} \), which slides down \( d = 1.35 \, \text{m} \) on an inclined plane with an angle \( \theta = 36.9^\circ \). The initial speed is zero since the block starts from rest. We need to find its final speed considering no friction.
02

Apply Energy Conservation Principle

The principle of conservation of energy states that the total mechanical energy of the system remains constant if only conservative forces are doing work. Therefore, the potential energy lost by the block is converted to kinetic energy. The equations are: Initial potential energy (PE) = gravitational potential energy, PE = \( m \cdot g \cdot h \), where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h \) is the vertical height the block descends.
03

Calculate Vertical Height

The height \( h \) can be found using the relationship between the distance along the plane and the vertical height: \( h = d \cdot \sin(\theta) = 1.35 \, \text{m} \cdot \sin(36.9^\circ) \). Calculate \( \sin(36.9^\circ) \) to find \( h \): \( h \approx 1.35 \, \text{m} \cdot 0.6 = 0.81 \, \text{m} \).
04

Compute Initial Potential Energy

The potential energy at the top is given by: \( PE = 2.00 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.81 \, \text{m} \). Calculate this to get the initial potential energy: \( PE = 15.88 \, \text{J} \).
05

Calculate Kinetic Energy at the Bottom

Since all the potential energy is converted into kinetic energy at the bottom (no non-conservative forces like friction are acting), the kinetic energy (KE) is equal to the initial potential energy: \( KE = 15.88 \, \text{J} \). The kinetic energy is given by: \( KE = \frac{1}{2} m v^2 \), where \( v \) is the final speed.
06

Solve for Final Speed

We solve for \( v \) using the equation: \( \frac{1}{2} \cdot 2.00 \, \text{kg} \cdot v^2 = 15.88 \, \text{J} \). Simplify to get: \( v^2 = 15.88 \). Take the square root to find \( v \): \( v \approx 3.99 \, \text{m/s} \).

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Key Concepts

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

Conservation of Energy
The principle of Conservation of Energy is a cornerstone in physics, allowing us to understand how energy is transferred and transformed within a system. When we talk about inclined plane problems, this principle becomes extremely useful. In a frictionless scenario, such as our block of ice sliding down the incline, no energy is lost due to external forces like friction. Instead, the potential energy at the start of the motion is completely transformed into kinetic energy by the end.

In mathematical terms, the initial mechanical energy, which is entirely potential in this context, equals the final mechanical energy, or kinetic energy:
  • Initial energy = Final energy
  • \[ m \cdot g \cdot h = \frac{1}{2} \cdot m \cdot v^2\]
The height \( h \) here is the component of the distance that accounts for the effect of gravity, given by the function \( \sin(\theta) \). Understanding this transformation simplifies finding unknown values, such as final speed, since we only need one form of energy to calculate the other once friction is ignored.
Kinetic Energy
Kinetic Energy is the energy a body possesses due to its motion. Imagine all the potential energy that a body had due to its position is now being fully utilized to set it in motion. For an object with mass moving at a speed \( v \), its kinetic energy is calculated with the formula:
  • \[KE = \frac{1}{2} \cdot m \cdot v^2\]
The beauty of kinetic energy in problems like our inclined plane exercise is that it captures the essence of motion provided by previously stored energy. When an object, like our ice block, slides down without friction, its increase in motion is precisely the energy it has gained by losing height.

In practical terms, once we calculated the initial potential energy (15.88 J in this case), we could directly relate it to the kinetic energy at the bottom of the incline. Using this relation, and knowing the mass of the block, allows us to easily solve for the final velocity, which completes the energy transformation process from potential to kinetic.
Potential Energy
Potential Energy, often related to position or state, serves as stored energy ready to be converted into motion or kinetic energy. In the context of an inclined plane, this energy is due to the gravitational force acting on the block of ice. As the block begins its journey down the slope, the potential energy is calculated with
  • \[PE = m \cdot g \cdot h\]
where \( m \) is the mass, \( g \) is the gravitational acceleration (9.81 \( \text{m/s}^2 \)), and \( h \) is the height derived from the inclined distance and angle. In our example, this involved determining \( h = d \cdot \sin(\theta) \), connecting how angles on an inclined plane influence the vertical potential energy change.

Once the block starts moving, this potential energy decreases as it's converted to kinetic energy. Recognizing the dominion of potential energy at the top of the slide, and the transition it undergoes, helps us visualize energy flow within the system and apply it to solve for subsequent values such as speed at the bottom.

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Most popular questions from this chapter

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 4.00-kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is \(\mu_k = 0.400\). You apply a constant force \(\overrightarrow{F}\) to the block. \(\overrightarrow{F}\) has magnitude \(F = 82.0\) N and is directed toward the wall. At the instant that the spring is compressed 80.0 cm, what are (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?

You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 2.00-kg block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 N is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched 0.400 m? (b) At that instant, what are the magnitude and direction of the acceleration of the block?

A boxed 10.0-kg computer monitor is dragged by friction 5.50 m upward along a conveyor belt inclined at an angle of 36.9\(^\circ\) above the horizontal. If the monitor's speed is a constant 2.10 cm/s, how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\overrightarrow{F} =(30 \, \mathrm{N})\hat{\imath} - (40 \, \mathrm{N})\hat{\jmath}\) to the cart as it undergoes a displacement \(\overrightarrow{s} = (-9.0 \, \mathrm{m})\hat{\imath} (3.0 \, \mathrm{m})\hat{\jmath}\). How much work does the force you apply do on the grocery cart?

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