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As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the \(10-\mathrm{kg}\) block which rests on the smooth surface and is subjected to a horizontal force of \(6 \mathrm{~N}\). If observer \(A\) is in a fixed frame \(x\), determine the final speed of the block if it has an initial speed of \(5 \mathrm{~m} / \mathrm{s}\) and travels \(10 \mathrm{~m}\), both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer \(B,\) attached to the \(x^{\prime}\) axis and moving at a constant velocity of \(2 \mathrm{~m} / \mathrm{s}\) relative to \(A\). Hint: The distance the block travels will first have to be computed for observer \(B\) before applying the principle of work and energy.

Short Answer

Expert verified
The final speed of the block will be the same as observed by both A and B. The principle of work and energy holds valid in different inertial reference frames.

Step by step solution

01

Calculate the Work Done

First, calculate the work done on the block. Work is the product of the horizontal force and the distance traveled. Use the formula \( W = F \cdot d \), where \( W \) is the work done, \( F \) is the horizontal force, and \( d \) is the distance. Plug in the given values: \( F = 6 \, \mathrm{N} \) and \( d = 10 \, \mathrm{m} \) to get the work done.
02

Calculate the Final Speed for Observer A

Apply the work and energy principle to find the block's final speed observed by A. The work done on the block is equal to the change in its kinetic energy. Hence, \( W = \Delta KE \), where \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \), \( m \) is the mass of the block, \( v_f \) is the final speed, and \( v_i \) is the initial speed. Solve this equation for \( v_f \) yielding the final speed as observed by A.
03

Calculate the Relative Distance for Observer B

Next, calculate the distance the block travels relative to observer B. B is moving at a constant speed of \( 2 \, \mathrm{m/s} \) in the same direction as the block. Thus, the distance \( d' \) the block travels relative to B is less than the distance it covers in the frame of A and can be calculated by the rule of relative velocity \( d' = d - v_B \cdot t \), where \( v_B \) is the speed of B and \( t \) is the time, which can be determined from the distance travelled and the initial speed of the block \( t = d / v_i \).
04

Calculate the Final Speed for Observer B

Use the work-energy principle again to find the block's final speed observed by B. Here, the work done is the same, but the distance \( d' \) and the initial speed \( v_i' = v_i - v_B \) are relative to B. This will yield the final speed as seen by B. Compare the final speeds obtained in Step 2 and Step 4.

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Key Concepts

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

Inertial Reference Frame
Understanding inertial reference frames is crucial when applying the principles of classical mechanics. An inertial reference frame is a perspective from which an observer witnesses events where no acceleration is occurring due to external forces. To put it simply, in such a frame, an object either remains at rest or moves at a constant velocity unless acted upon by an external force. It is this predictability that allows us to apply the laws of physics consistently.

Using our block on the smooth surface as an example, observer A in a fixed frame observes the block accelerating due to a constant horizontal force. Since there is no unbalanced force in the frame of A, it can be considered an inertial reference frame. On the other hand, observer B is moving at a constant velocity and is not accelerating, indicating that B is also in an inertial reference frame. The principle of work and energy holds true for both observers because they are both in inertial frames, despite their relative motion to each other.
Kinetic Energy
Kinetic energy, often abbreviated as KE, is a form of energy associated with an object's motion. It depends on the mass of the object and the square of its velocity. We can express it quantitatively as
\(\text{KE} = \frac{1}{2} m v^2\)
, where
\(m\) is the mass and \(v\) is the velocity of the object. When a constant force is applied to a block, like in our example, the block's kinetic energy changes.

The principle of work and energy states that the work done on the object results in a change in its kinetic energy. If we know the work done on the block and its initial kinetic energy, we can determine its final kinetic energy and consequently the final speed. This relationship shows that energy can be transformed from one form to another, but the total amount of energy remains constant within an inertial reference frame.
Relative Velocity
Relative velocity is the velocity of one object as observed from another moving object. In many physical scenarios, the observer's frame of reference significantly affects how the motion is perceived. In our step-by-step exercise, observer B's frame of reference has a velocity of
\(2 \mathrm{~m} / \mathrm{s}\)
relative to observer A's stationary frame.

The relative velocity of the block as observed by B would be its velocity in A's frame minus B's own velocity. In mathematical terms, if \(v_i\) is the initial velocity of the block in the fixed frame and \(v_B\) is the velocity of observer B relative to A, then the initial velocity of the block relative to B is \(v_i' = v_i - v_B\). This must be accounted for when calculating the work done and energy changes from B's perspective. This principle is key when dealing with problems in mechanics, as it ensures the correct frame of reference is used to analyze the motion relative to the observer.

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Most popular questions from this chapter

The catapulting mechanism is used to propel the \(10-\mathrm{kg}\) slider \(A\) to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to \(\operatorname{rod} B C\) rapidly to the left by means of a piston \(P\). If the piston applies a constant force \(F=20 \mathrm{kN}\) to \(\operatorname{rod} B C\) such that it moves it \(0.2 \mathrm{~m},\) determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and \(\operatorname{rod} B C\).

The force \(\mathbf{F}\), acting in a constant direction on the 20 -kg block, has a magnitude which varies with the position \(s\) of the block. Determine how far the block must slide before its velocity becomes \(15 \mathrm{~m} / \mathrm{s}\). When \(s=0\) the block is moving to the right at \(v=6 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the block and surface is \(\mu_{k}=0.3\).

If the mass of the earth is \(M_{e}\), show that the gravitational potential energy of a body of mass \(m\) located a distance \(r\) from the center of the earth is \(V_{g}=-G M_{e} m / r\). Recall that the gravitational force acting between the earth and the body is \(F=G\left(M_{e} m / r^{2}\right),\) Eq. 13-1. For the calculation, locate the datum at \(r \rightarrow \infty\). Also, prove that \(F\) is a conservative force.

The rocket sled has a mass of \(4 \mathrm{Mg}\) and travels from rest along the horizontal track for which the coefficient of kinetic friction is \(\mu_{k}=0.20 .\) If the engine provides a constant thrust \(T=150 \mathrm{kN},\) determine the power output of the engine as a function of time. Neglect the loss of fuel mass and air resistance.

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's velocity is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\) directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3,\) determine the speed at which each crate slides off the ramp at \(B\). Assume that no tipping occurs.

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