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(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's velocity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W},\) what is the velocity of the object?

Short Answer

Expert verified
(a) 28 W; (b) Velocity is 6 m/s in the j-direction

Step by step solution

01

Understand the Problem

We need to calculate the instantaneous rate at which a force does work, which is given by the power due to the force. The power is the dot product of the force vector and the velocity vector. Given two scenarios, we first solve for one and then use the power to solve for the unknown velocity in the other.
02

Calculate Power for First Scenario

Power is calculated with the formula: \( P = \vec{F} \cdot \vec{v} \). Given \( \vec{F} = (4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}} \) and \( \vec{v} = -(2.0 \mathrm{~m/s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m/s}) \hat{\mathrm{k}} \), calculate the dot product: \( P = (4)(-2) + (9)(4) = -8 + 36 = 28 \). Thus, \( P = 28 \mathrm{~W} \).
03

Identify Variables for Second Scenario

In the second scenario, the force \( \vec{F} \) remains the same, but the velocity has only a \( y \)-component, so let it be \( v_y \). The power is given as \( -12 \mathrm{~W} \). We can use the power formula again to find \( v_y \).
04

Solve for Velocity in Second Scenario

The power formula in the second scenario is \( P = \vec{F} \cdot (v_y \hat{\mathrm{j}}) = (-2.0 \mathrm{~N})(v_y) = -12 \mathrm{~W} \). Solving for \( v_y \), we have \( -2v_y = -12 \). So, \( v_y = 6 \mathrm{~m/s} \).

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Key Concepts

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

Force Vector
A force vector describes the influence that produces a change in the motion of an object. It is a vector quantity, meaning it has both magnitude and direction. In physics problems, forces are often expressed in terms of their components along the coordinate axes, such as the x, y, and z axes.

For the given exercise, the force vector is defined as \( \vec{F}=(4.0 \,\text{N}) \hat{\text{i}}-(2.0 \,\text{N}) \hat{\text{j}}+(9.0 \,\text{N}) \hat{\text{k}} \).
This means the force has:
  • A component of 4.0 N in the x-direction
  • A component of -2.0 N in the y-direction
  • A component of 9.0 N in the z-direction
Each component is associated with its axis-specific unit vector \( \hat{\text{i}} \), \( \hat{\text{j}} \), or \( \hat{\text{k}} \), indicating the direction of each component.
Understanding the components of a force vector is crucial as it allows us to consider the force's influence along each individual direction.
Velocity Vector
A velocity vector represents the speed and direction of an object’s motion. Like the force vector, it is also expressed in terms of its components along coordinate axes.

In the original problem, the velocity vector is given as \( \vec{v}=-(2.0 \,\text{m/s}) \hat{\text{i}}+(4.0 \,\text{m/s}) \hat{\text{k}} \).
This indicates:
  • The object is moving at 2.0 m/s in the negative x-direction
  • The object is moving at 4.0 m/s in the positive z-direction
Note that there is no y-component specified, suggesting no movement in the y-direction in the first scenario. Velocity vectors are essential in determining how different forces affect an object's motion, as they factor into the calculation of instantaneous power.
Dot Product
The dot product is a method to multiply two vectors, producing a scalar quantity. It is crucial in calculating work and power in physics because it inherently considers the direction of vectors.

For two vectors \( \vec{A} \) and \( \vec{B} \), where \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \) and \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \), the dot product is calculated as:\[\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]
In the problem at hand, the force and velocity vectors are used to find power:
  • \( \vec{F} = (4.0 \text{N}) \hat{i} - (2.0 \text{N}) \hat{j} + (9.0 \text{N}) \hat{k} \)
  • \( \vec{v} = -(2.0 \text{m/s}) \hat{i} + (4.0 \text{m/s}) \hat{k} \)
The dot product for the power calculation is:\[P = (4 \times -2) + (9 \times 4) = -8 + 36 = 28 \]Understanding the dot product helps us compute how aligned two vectors are, which is essential in understanding work done or power transferred when vectors are not perfectly aligned.
Power Calculation
Power, in the context of physics, is the rate at which work is done or energy is transferred over time. When dealing with forces and motion, instantaneous power can be calculated using the dot product of the force and velocity vectors. This implies that power depends on how much of the force is acting in the direction of velocity.

In the first scenario of the problem, the power calculation is \[P = \vec{F} \cdot \vec{v} = 28 \, \text{W}\]This indicates the force is doing positive work on the object, leading to a power output of 28 Watts.

In the second scenario, we know the force is constant and power is given as -12 W with velocity having only a y-component, \( v_y \). Using the formula:\[P = -2 \cdot v_y = -12 \]Solving this gives:\[v_y = 6 \, \text{m/s} \] Power is an important concept as it gauges how effectively force displaces an object. Understanding power calculation allows one to infer energy dynamics in physical systems.

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Most popular questions from this chapter

A force of \(5.0 \mathrm{~N}\) acts on a \(15 \mathrm{~kg}\) body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN},\) was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm},\) how much work did her force do on the car?

On August \(10,1972,\) a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite's mass was about \(4 \times 10^{6} \mathrm{~kg}\) : its speed was about \(15 \mathrm{~km} / \mathrm{s}\). Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of \(\mathrm{TNT}\), which is \(4.2 \times 10^{15} \mathrm{~J}\). (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

If a car of mass \(1200 \mathrm{~kg}\) is moving along a highway at \(120 \mathrm{~km} / \mathrm{h}\) what is the car's kinetic energy as determined by someone standing alongside the highway?

A fully loaded, slow-moving freight elevator has a cab with a total mass of \(1200 \mathrm{~kg}\), which is required to travel upward \(54 \mathrm{~m}\) in \(3.0 \mathrm{~min},\) starting and ending at rest. The elevator's counterweight has a mass of only \(950 \mathrm{~kg}\), and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

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