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A \(12.0 \mathrm{~N}\) force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement \(\vec{d}=(2.00 \hat{i}-4.00 \mathrm{j}+3.00 \mathrm{k}) \mathrm{m} .\) What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) \(+30.0 \mathrm{~J}\) and (b) \(-30.0 \mathrm{~J} ?\)

Short Answer

Expert verified
For (a), the angle is \(62.39^\circ\), and for (b), it is \(117.61^\circ\).

Step by step solution

01

Understand the Relationship Between Work and Energy

We need to find the angle between the force and the displacement when given specific changes in kinetic energy. In physics, the work-energy principle states that the work done by an external force on an object equals the change in kinetic energy of that object.
02

Formula for Work Done by a Force

The work \(W\) done by a force \(\vec{F}\) through a displacement \(\vec{d}\) is given by the dot product: \(W = \vec{F} \cdot \vec{d} = Fd \cos \theta\), where \(\theta\) is the angle between the force and the displacement, \(F\) is the magnitude of the force, and \(d\) is the magnitude of the displacement.
03

Calculate the Magnitude of Displacement

To find the magnitudes, calculate the displacement \(|\vec{d}|\) using the formula: \(|\vec{d}| = \sqrt{(2.00)^2+(-4.00)^2+(3.00)^2}\). Evaluate to get \(|\vec{d}| = \sqrt{4 + 16 + 9} = \sqrt{29} \approx 5.39 \mathrm{~m}\).
04

Use the Work for Kinetic Energy Change (Case a)

The condition for case (a) is a change in kinetic energy of \(+30.0 \mathrm{~J}\). Set the work done equal to the change in energy: \(12 \times 5.39 \times \cos \theta = 30\).
05

Solve for \(\theta\) (Case a)

Solve \(64.68 \cos \theta = 30\). Then, \(\cos \theta = \frac{30}{64.68} \approx 0.464\). Use the inverse cosine function to get \(\theta = \cos^{-1}(0.464) \approx 62.39^\circ\).
06

Work for Kinetic Energy Change (Case b)

For case (b), the change in kinetic energy is \(-30.0 \mathrm{~J}\). Set up the equation \(12 \times 5.39 \times \cos \theta = -30\).
07

Solve for \(\theta\) (Case b)

Solving \(64.68 \cos \theta = -30\) gives \(\cos \theta = \frac{-30}{64.68} \approx -0.464\). Thus, \(\theta = \cos^{-1}(-0.464) \approx 117.61^\circ\).

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Key Concepts

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

Dot Product
The dot product is a mathematical operation that offers a way to multiply two vectors. This operation is essential in physics for calculating work done by a force. When you have a force vector \(\vec{F}\) and a displacement vector \(\vec{d}\), the dot product \(\vec{F} \cdot \vec{d}\) gives a scalar quantity. This scalar represents the component of the force that is effective in doing work along the direction of the displacement.
The formula for the dot product in terms of magnitudes and the angle \(\theta\) between the vectors is:
  • \(\vec{F} \cdot \vec{d} = Fd \cos \theta\)
Where:
  • \(F\) is the magnitude of the force
  • \(d\) is the magnitude of the displacement
  • \(\cos \theta\) reflects how aligned the vectors are
In our exercise, the dot product helps determine how the force influences change in kinetic energy. It calculates whether the force increases, decreases, or does not change the energy, based on the angle between the force and displacement.
Angle between Force and Displacement
In physics, the angle \(\theta\) between the force applied and the displacement vector is crucial for understanding how effective the force is in doing work on an object. This angle affects the amount of work done since work requires that part of the force acts in the direction of the displacement.
When calculating the angle from the known change in kinetic energy, we use the relationship:
  • \(W = Fd \cos \theta \)
Here, \(W\) represents the work done, equivalent to the change in kinetic energy.
In this context:
  • If \(\theta = 0^\circ\): Force and displacement are perfectly aligned, maximizing work.
  • If \(\theta = 90^\circ\): Force is perpendicular to displacement, resulting in zero work done.
  • If \(\theta = 180^\circ\): Force is opposite to displacement, doing negative work, thus decreasing kinetic energy.
The exercise gives examples for positive and negative changes in kinetic energy to show how variations in \(\theta\) lead to different outcomes.
Change in Kinetic Energy
The change in kinetic energy is a significant concept in mechanics, epitomizing how a force affects the motion of a particle. According to the work-energy principle, any work done on a particle results in a change in its kinetic energy. This is expressed mathematically as \(\Delta KE = W\), where \(\Delta KE\) is the change in kinetic energy.
The exercise scenarios provide two cases, one with a positive kinetic energy change \((+30.0 \, \text{J})\) indicating that the force added energy resulting in faster motion, and another with a negative change \((-30.0 \, \text{J})\) illustrating energy loss, slowing down the object.
In practical terms:
  • The size of this energy change depends directly on the angle \(\theta\) as it regulates the work done.
  • Positive energy change improves particle speed, aligning \(\theta\) closer to \(0^\circ\).
  • Negative energy change implies force opposes displacement, with \(\theta\) closer to \(180^\circ\).
Analyzing changes in kinetic energy through equations provided in exercises helps understand how energies interact due to external forces, offering deep insights into motion dynamics.

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Most popular questions from this chapter

If a ski lift raises 100 passengers averaging \(660 \mathrm{~N}\) in weight to a height of \(150 \mathrm{~m}\) in \(60.0 \mathrm{~s}\), at constant speed, what average power is required of the force making the lift?

A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

If a Saturn \(\mathrm{V}\) rocket with an Apollo spacecraft attached had a combined mass of \(2.9 \times 10^{5} \mathrm{~kg}\) and reached a speed of \(11.2 \mathrm{~km} / \mathrm{s}\), how much kinetic energy would it then have?

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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