/*! 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 74 A particle moves with a velocity... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle moves with a velocity \(5 \hat{i}-3 \hat{j}+6 \hat{k} \mathrm{~m} / \mathrm{s}\) under the influenec of a constant force \(\vec{F}-10 \hat{i}+10 \hat{j}+20 \hat{k} \mathrm{~N}\). The instantaneous power applied to the particle is (a) \(200 \mathrm{~J} / \mathrm{s}\) (b) \(40 \mathrm{~J} / \mathrm{s}\) (c) \(140 \mathrm{~J} / \mathrm{s}\) (d) \(170 \mathrm{~J} / \mathrm{s}\)

Short Answer

Expert verified
The correct answer is (b) 40 J/s.

Step by step solution

01

Understand the Formula for Power

The instantaneous power applied by a force on a moving particle is given by the dot product of the force vector \(\vec{F}\) and the velocity vector \(\vec{v}\), i.e., \( P = \vec{F} \cdot \vec{v} \). We will use this relationship to find the instantaneous power.
02

Identify the Vectors

The force vector is given as \( \vec{F} = -10 \hat{i} + 10 \hat{j} + 20 \hat{k} \) N and the velocity vector is \( \vec{v} = 5 \hat{i} - 3 \hat{j} + 6 \hat{k} \) m/s.
03

Compute Dot Product

The dot product of \(\vec{F}\) and \(\vec{v}\) is given by \[ \vec{F} \cdot \vec{v} = (-10)(5) + (10)(-3) + (20)(6) \] The calculation is as follows:- \((-10) \cdot 5 = -50\)- \(10 \cdot (-3) = -30\)- \(20 \cdot 6 = 120\) Adding these results:\[ -50 - 30 + 120 = 40 \]
04

Conclusion

The instantaneous power applied to the particle is \( 40 \) J/s. Therefore, the correct choice is (b) \( 40 \mathrm{~J} / \mathrm{s} \).

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Key Concepts

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

Dot Product Basics
When dealing with vector quantities that have both magnitude and direction, the dot product is a mathematical operation that helps. Simply put, it multiplies corresponding components of two vectors and sums them up to produce a single scalar value. For example, if you have two vectors, \( \vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), the dot product \( \vec{A} \cdot \vec{B} \) is given by:
  • \( a_1b_1 + a_2b_2 + a_3b_3 \)
This results in a number (a scalar) which can describe, in this context, the power delivered by a force at any given instant. Using the dot product in physics is essential when you want to determine how a force vector influences motion defined by a velocity vector.
Understanding the Force Vector
A force vector represents a force's magnitude and direction, usually depicted in Newtons in physics problems. It's a fundamental aspect of understanding how objects move or react under certain forces. In our case, the force vector is \( \vec{F} = -10 \hat{i} + 10 \hat{j} + 20 \hat{k} \text{ N} \). Each component illustrates the force along the respective axes:
  • \( -10 \text{ N} \) along the \( x \)-axis
  • \( 10 \text{ N} \) along the \( y \)-axis
  • \( 20 \text{ N} \) along the \( z \)-axis
Recognizing how force vectors work helps in visualizing the complete force being exerted on a particle and is critical for computing effects like power and work.
Decoding the Velocity Vector
Velocity vectors describe an object's speed along with its direction of motion. The components \( \vec{v} = 5 \hat{i} - 3 \hat{j} + 6 \hat{k} \text{ m/s} \) identify the speed in each coordinate direction:
  • \( 5 \text{ m/s} \) along the \( x \)-axis
  • \( -3 \text{ m/s} \) along the \( y \)-axis
  • \( 6 \text{ m/s} \) along the \( z \)-axis
Understanding the velocity vector is crucial for predicting where an object will be next and how forces will influence its movement. When combined with vectors like force, it provides insights into momentum, acceleration, and power.
Applying the Power Formula
The instantaneous power formula provides a means to determine the power applied at a specific instant. Power is calculated using the formula \( P = \vec{F} \cdot \vec{v} \), which represents the dot product of the force and velocity vectors. It shows how effectively a force is being used to cause movement. Calculating this involves multiplying and summing the vector components:
  • Find the product of \( x \)-components
  • Find the product of \( y \)-components
  • Find the product of \( z \)-components
  • Add these products together
In this example, the power calculated was \( 40 \text{ J/s} \), indicating how force contributes instantaneously to a particle’s kinetic energy change.
Performing Vector Calculations
Vector calculations fundamentally involve performing arithmetic operations on these mathematical entities, often to describe physical phenomena like motion. When you calculate vectors:
  • Align corresponding components for operations
  • Multiply or add as necessary based on the vector operation type (e.g., addition, dot product)
  • Consider both magnitude and direction to give meaning to the results
These calculations empower us to analyze and forecast behavior in systems like mechanical structures or flowing fluids. Mastery of vector math enables students and professionals alike to solve complex problems and appreciate the intricate dance of forces and motion.

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Most popular questions from this chapter

A small disc of mass \(m\) slides down a smooth hill of height \(h\) without any initial velocity and lands upon a plank of mass \(M\) lying along the horizontal plane at the base of the hill. Due to friction between the disc and the plank the disc slows down till the two move together as a single piece with a certain common velocity. Find the work done by the force of friction in the process. Solution Since mass \(m\) falls through hcight \(h\) just before landing on the plank of mass \(M\), so velocity of at this moment is \(v-\sqrt{2 g h}\). Now, the frictional force betwcen \(M\) and \(m\) is the internal force the system so momentum of system remains conserved. I lence, \(m v-(M \mid m) v^{\prime}\) or \(v^{\prime}-\frac{m v}{M+m}\) or \(v^{\prime}-\frac{m \sqrt{2 g h}}{M+m}\) So, initial kinetic cnergy of the system \(K_{i}-\frac{1}{2} m v^{2} \quad\) or \(\quad K_{i}-m g h\) \(\ldots(2)\) and linal kinctic cnergy of the system \(K_{f}-\frac{1}{2}(M+m) v^{\prime 2}-\frac{1}{2}(M+m) \frac{m^{2} \times 2 g h}{(M+m)^{2}} \quad \Rightarrow K_{f}-\frac{m^{2} g h}{M+m} \quad \ldots(3)\) So, change of cnergy \(\Delta K-K_{f} K_{t} \Rightarrow \Delta K-\frac{m M g h}{M \mid m}\) Hence, work-done by friction \(W-\Delta K \quad \therefore W--\frac{m M g h}{M \mathrm{I} m}\)

Statement-1: In a circular motion, the force must be dirccied perpendicular to the velocily all the time. Statement-2: \(\Lambda\) centripetal forec is required to provide the centripelal acecleration in a circular motion.

No work is done by a force on an object if (a) the force is always perpendicular to its velocity. (b) the force is always perpendicular to its acceleration. (c) the object is stationary but the point of application of the force moves on the object. (d) the object moves in such a way that the point of application of the force remains fixed.

The potential energy of a particle of mass \(1 \mathrm{~kg}\) is, \(U=10+(x-2)^{2} .\) Herc, \(U\) is in joules and \(x\) in metres on the positive \(x\) -axis. Particle travels upto \(x=+6 m\). Choose the correct statement. (a) On negative \(x\) -axis particle travels upto \(x=-2 \mathrm{~m}\) (b) The maximum kinetic energy of the particle is \(16 \mathrm{~J}\) (c) Both (a) and (b) are correct (d) Both (a) and (b) are wrong

In certain position \(X\), kinetic energy of a particle is \(25 \mathrm{~J}\) and potential energy is \(-10 \mathrm{~J} .\) In another position \(Y\), kinetic energy of the particle is \(95 \mathrm{~J}\) and the potential energy is \(-25 \mathrm{~J}\). During the displacement of the particle from \(X\) to \(Y\) (a) Net work done by all the forces is \(-70 \mathrm{~J}\). (b) Work done by the conservative forces is \(-15 \mathrm{~J}\). (c) Work done by all the forces besides the conservative forces is \(55 \mathrm{~J}\). (d) Work done by the conservative forces equals work done by the non- conservative forces.

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