/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 A block is sent sliding down a f... [FREE SOLUTION] | 91Ó°ÊÓ

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A block is sent sliding down a frictionless ramp. Its speeds at points \(A\) and \(B\) are \(2.00 \mathrm{~m} / \mathrm{s}\) and \(2.60 \mathrm{~m} / \mathrm{s}\), respectively. Next, it is again sent sliding down the ramp, but this time its speed at point \(A\) is \(4.00 \mathrm{~m} / \mathrm{s}\). What then is its speed at point \(B ?\)

Short Answer

Expert verified
The speed at point B is 5.00 m/s.

Step by step solution

01

Understand the Problem

The problem involves a block sliding down a frictionless ramp, which means that mechanical energy is conserved. We are given two scenarios for the block's motion: first with speeds at points A and B as 2.00 m/s and 2.60 m/s, and then with a new speed at point A to determine the speed at point B.
02

Apply Energy Conservation for First Scenario

Use the conservation of mechanical energy. For frictionless motion, \[ \frac{1}{2}m v_A^2 + mgh_A = \frac{1}{2}m v_B^2 + mgh_B \]Given: Speed at A, \(v_A = 2.00 \text{ m/s}\), speed at B, \(v_B = 2.60 \text{ m/s}\). Since there is no height change due to frictionless condition,\[ \frac{1}{2}(2.00)^2 = \frac{1}{2}(2.60)^2 \]
03

Determine Height Change

The potential energy change is converted to kinetic energy. Calculate height change:\[ v_B^2 - v_A^2 = 2g(h_A - h_B) \]Using known speeds:\[ (2.60)^2 - (2.00)^2 = 2gh \]Solve for h (height difference).
04

Apply Energy Conservation for Second Scenario

Again using energy conservation:\[ \frac{1}{2}m v_{A'}^2 + mgh_{A'} = \frac{1}{2}m v_{B'}^2 + mgh_{B'} \]New speed at A, \(v_{A'} = 4.00 \text{ m/s}\). Since height change equivalent,\[ \frac{1}{2}(4.00)^2 = \frac{1}{2}v_{B'}^2 + g(h_{A} - h_{B}) \]Substitute h from Step 3.
05

Solve for speed at B in second scenario

Finally:Using the earlier calculated height,Solve for \(v_{B'}\):\[ (4.00)^2 + 2gh = v_{B'}^2 \]This will give us the new speed at point B.
06

Calculate the Result

Calculate the value of the new speed \(v_{B'}\) using the equation:\[ v_{B'} = \sqrt{16 + (2.60)^2 - (2.00)^2} \]Solve and simplify the calculation to find the final speed at point B.

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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 a form of energy associated with the motion of an object. Whenever an object is moving, it possesses kinetic energy. It is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]where \(m\) is the mass of the object and \(v\) is its velocity. This formula tells us that the kinetic energy of an object is directly proportional to its mass and the square of its velocity. As such, even small changes in the speed of an object can lead to significant changes in its kinetic energy. This is particularly relevant in our exercise where small changes in speed from point A to B illustrate how kinetic energy varies with velocity.

It’s important to understand that kinetic energy is a scalar quantity, meaning it only has magnitude and no direction. This is different from forces and velocity, which are vector quantities. When a block slides down a ramp, like in this problem, its kinetic energy increases as the speed of the block increases.
Potential Energy
Potential energy refers to the energy stored within a system, due to an object's position or state. In the context of our exercise, we discuss gravitational potential energy, which is the energy stored as a result of an object's height relative to some reference point. This energy is expressed by:\[ PE = mgh \]Here, \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height above the reference point. As the block slides down the ramp, its height decreases, causing its potential energy to decrease as well.

The conservation of mechanical energy tells us that the total mechanical energy (sum of kinetic and potential energy) in a frictionless system remains constant. Hence, the decrease in potential energy as the block descends is converted entirely into an increase in kinetic energy. Understanding this transfer between potential and kinetic energy is fundamental to solving the problem correctly and illustrates how energy conservation operates in frictionless environments.
Frictionless Motion
Frictionless motion occurs when there are no forces opposing the movement. This simplifies problems significantly because friction can complicate the energy calculations. In a frictionless scenario, mechanical energy is conserved, meaning the total energy remains constant.

This principle is crucial for solving our problem, as it allows us to use conservation of energy to analyze the block's motion without worrying about energy loss to heat or sound, which are typical with friction. The beauty of frictionless motion lies in the simple transformation between kinetic and potential energy.

  • Kinetic energy increase as the block moves faster down the ramp.
  • Potential energy decrease as the block loses height.
  • Total mechanical energy remains constant.
In the absence of friction, the energy calculations become straightforward, allowing us to predict the block's speed at various points along the ramp accurately by using initial conditions and the laws of physics governing energy conservation.

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Most popular questions from this chapter

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain click-beetle jump shows that a beetle of mass \(m=4.0 \times 10^{-6} \mathrm{~kg}\) moved directly upward by \(0.77 \mathrm{~mm}\) during the launch and then to a maximum height of \(h=0.30 \mathrm{~m}\). During the launch, what are the average magnitudes of (a) the external force on the beetle's back from the floor and (b) the acceleration of the beetle in terms of \(g\) ?

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N}\), the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\) with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\).

A block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m}\). When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm} ?\)

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?(\mathrm{~d})\) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

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