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You are applying a constant horizontal force \(\overrightarrow{F} = (-8.00\mathrm{N})\hat{\imath} + (3.00\mathrm{N})\hat{\jmath}\) to a crate that is sliding on a factory floor. At the instant that the velocity of the crate is \(\overrightarrow{\upsilon} = (3.20\mathrm{m/s})\hat{\imath} + (2.20 \mathrm{m/s})\hat{\jmath}\), what is the instantaneous power supplied by this force?

Short Answer

Expert verified
The instantaneous power is \(-19.00\, \mathrm{W}\).

Step by step solution

01

Understand the Concept of Power

Power is the rate at which work is done, and in this context, it can be calculated using the dot product of the force vector and the velocity vector. The formula for instantaneous power is given by \( P = \overrightarrow{F} \cdot \overrightarrow{v} \), where \( \overrightarrow{F} \) is the force and \( \overrightarrow{v} \) is the velocity.
02

Apply the Dot Product

Compute the dot product of the force vector \((-8.00\mathrm{N})\hat{\imath} + (3.00\mathrm{N})\hat{\jmath}\) and the velocity vector \((3.20\mathrm{m/s})\hat{\imath} + (2.20 \mathrm{m/s})\hat{\jmath}\). The dot product is calculated as follows: \( (-8.00 \times 3.20) + (3.00 \times 2.20) \).
03

Calculate Each Component

Find the result of each multiplication: \((-8.00) \times (3.20) = -25.60\) and \((3.00) \times (2.20) = 6.60\).
04

Sum the Components

Add the results of the components: \(-25.60 + 6.60 = -19.00\). This sum represents the instantaneous power.
05

Interpret the Result

The result \(-19.00 \mathrm{W}\) is the instantaneous power supplied by the force. The negative sign indicates that the force is acting opposite to the direction of the velocity, possibly doing work on other systems or overcoming some form of resistance.

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Key Concepts

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

Understanding the Dot Product
The dot product is a way to multiply two vectors, resulting in a scalar. It quantifies how much one vector goes in the direction of another. To calculate it, you multiply corresponding components and then sum them up. Consider vectors \( \overrightarrow{F} = (-8.00 \mathrm{N})\hat{\imath} + (3.00 \mathrm{N})\hat{\jmath} \) and \( \overrightarrow{v} = (3.20 \mathrm{m/s})\hat{\imath} + (2.20 \mathrm{m/s})\hat{\jmath} \). The dot product is calculated like this:
  • Multiply the \( \hat{\imath} \)-components: \((-8.00) \times (3.20) = -25.60\)
  • Multiply the \( \hat{\jmath} \)-components: \((3.00) \times (2.20) = 6.60\)
After finding the products, add them: \(-25.60 + 6.60 = -19.00\). Thus, the dot product here is \(-19.00\).
This result, a single number, tells us about the directional alignment and magnitude of interaction between the vectors.
The Role of the Force Vector
A force vector represents both the magnitude and direction of a force. In our example, \( \overrightarrow{F} = (-8.00 \mathrm{N})\hat{\imath} + (3.00 \mathrm{N})\hat{\jmath} \). Each component defines how much force is applied along a specific axis:
  • -8.00 N in the direction of \( \hat{\imath} \), meaning force is applied left.
  • 3.00 N in the direction of \( \hat{\jmath} \), indicating upward force.
These components impact how the crate moves. A vector allows us to consider forces separately along each axis, simplifying complex motion into understandable parts. Knowing the vector helps analyze forces in physics problems.
Understanding the Velocity Vector
The velocity vector indicates both the speed and direction of an object's motion. Here, it is given by \( \overrightarrow{v} = (3.20 \mathrm{m/s})\hat{\imath} + (2.20 \mathrm{m/s})\hat{\jmath} \). Each part describes movement:
  • 3.20 m/s in \( \hat{\imath} \), indicating rightward movement.
  • 2.20 m/s in \( \hat{\jmath} \), indicating upward movement.
The vector reflects how the crate travels over time. By using vectors, changes in speed or direction can be easily composed or decomposed into simpler parts. As with force, this approach reveals the complete picture of motion in a clear mathematical form.
Rate of Work and Instantaneous Power
Instantaneous power is the power at a specific moment, embodying the concept of the rate at which work is done. When a force acts on an object in motion, like the crate, power can be derived from the dot product of the force and velocity vectors: \[ P = \overrightarrow{F} \cdot \overrightarrow{v} = -19.00 \text{ W} \]This tells us how "quickly" force is doing work over time. Here, the negative result \(-19.00\) W indicates that the force's direction predominantly counters the crate's motion, such as overcoming friction or other forces. Thus, instantaneous power not only measures energy transformation but also gives insight into motion dynamics and resistance.

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Most popular questions from this chapter

On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.0 N and direction 37.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does this force do during a displacement of the pig that is (a) \(\overrightarrow{s} = (5.00 \, \mathrm{m})\hat{\imath}\); (b) \(\overrightarrow{s} = - (6.00 \, \mathrm{m})\hat{\jmath}\); (c) \(\overrightarrow{s} = - (2.00 \, \mathrm{m})\hat{\jmath} + (4.00 \, \mathrm{m}) \hat{\jmath}\)?

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work\(-\)energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

The spring of a spring gun has force constant \(k = 400\) N/m and negligible mass. The spring is compressed 6.00 cm, and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so that the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

(a) In the Bohr model of the atom, the ground state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 m, how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a 30-kg child could run fast enough to have 100 J of kinetic energy?

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