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When a body moves in a circle, the work done by the centripetal force is always (A) \(>0\) (B) \(<0\) (C) Zero (D) None of these

Short Answer

Expert verified
The work done by the centripetal force on a body moving in a circle is always zero because the centripetal force is always perpendicular to the displacement vector. Thus, the angle between the force and displacement vectors is 90 degrees, and the cosine of 90 degrees is 0. Therefore, the correct answer is (C) Zero.

Step by step solution

01

Recalling the centripetal force formula

Centripetal force is calculated using the formula: \(F_c = \frac{mv^2}{r}\), where \(F_c\) is the centripetal force, \(m\) is the mass of the object, \(v\) is the velocity of the object, and \(r\) is the radius of the circle.
02

Understanding the direction of centripetal force

The direction of the centripetal force is always towards the center of the circular path. Hence, by definition, it is always perpendicular to the velocity vector of the object.
03

Calculating the work done by the centripetal force

To find the work done by the centripetal force, we need to calculate the dot product of the force vector and the displacement vector (remember that work done is equal to W = F * d * cosθ). Since the centripetal force is always directed towards the center of the circular path, it is always perpendicular to the velocity vector and the tangent of the path (which is the direction of the displacement vector). Thus, the angle between the centripetal force vector and the displacement vector is always 90 degrees. The cosine of 90 degrees is 0. Therefore, the work done by the centripetal force on a body moving in a circle is: \[W = F_c * d * \cos(90°) = F_c * d * 0 = 0\]
04

Identifying the correct answer

The work done by the centripetal force on a body moving in a circle is always zero, which means the correct answer for this exercise is (C) Zero.

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Key Concepts

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

Circular Motion
Circular motion is a fundamental concept in physics that describes the movement of an object along a circular path. It's critical for students, particularly those preparing for exams like the JEE MAIN, to understand the intricacies of circular motion, as it occurs frequently in nature and technology.

When discussing circular motion, it's essential to consider the forces that are acting on an object. The most notable force in this context is the centripetal force, which acts towards the center of the circle, keeping the object moving in a curved path rather than in a straight line. According to the formula \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius, we see how velocity and radius influence the necessary force to maintain circular motion.

In addition to the centripetal force, other forces might be present, such as tension in a string, gravitational force, or frictional force, depending on the context of the motion. The role of each force must be analyzed to fully understand the motion of the body.
Work-Energy Theorem
The work-energy theorem is a vital principle in physics which connects the work done by forces on an object to changes in kinetic energy. This theorem plays a pivotal role in solving problems related to energy and work for exams like the JEE MAIN, making it a core concept for aspiring engineers and physicists.

The theorem states that the work done by the total force acting on a particle is equal to the change in its kinetic energy. Mathematically, this is represented as \( W = ΔKE \), where \( W \) is the work done, and \( ΔKE \) represents the change in kinetic energy. It's important to note that work and energy are scalar quantities — they have magnitude but no direction.

An interesting aspect of the work-energy theorem in the context of circular motion is that even though a centripetal force is needed to maintain the motion, it does not do work on the body moving in a circular path. This is because the force acts perpendicular to the direction of motion. Hence, there's no change in the kinetic energy of the object due to the centripetal force, reinforcing the fact that it does zero work, as observed in the textbook solution.
Physics for JEE MAIN
Physics is a significant segment of the JEE MAIN syllabus, which is a standardized test in India for students aspiring to enter engineering colleges. Preparing for physics in the JEE MAIN involves a deep understanding of concepts, problem-solving abilities, and application of principles like those related to circular motion and work-energy theorem.

Students are often tested on their ability to apply fundamental concepts to solve complex problems efficiently. For instance, the problem related to the work done by centripetal force combines the concepts of forces, motion, and work-energy theorem. The aspirants are expected to not only memorize the formulas but also to understand the principles behind them, such as the relationship between force vectors and work done.

Furthermore, JEE MAIN aspirants need to develop critical thinking skills to identify the correct approach to problems. They should practice various problems to familiarize themselves with different scenarios, enhancing their analytical skills and their ability to tackle challenging questions under exam conditions.

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Most popular questions from this chapter

A block of mass \(0.5 \mathrm{~kg}\) is kept in an elevator moving down with an acceleration \(2 \mathrm{~m} / \mathrm{s}^{2}\). Find the magnitude work done (in Joule) by the normal contact force on the block in first second. Initially system is at rest \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

Two identical beads of \(m=100 \mathrm{~g}\) are connected by an inextensible massless string that can slide along the two arms \(A C\) and \(B C\) of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of \(0.1 \mathrm{~m}\) is \(16 x \times 10^{-3} \mathrm{~J}\). Find the value of \(x .\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

A wire suspended vertically from one of its ends is stretched by attaching a weight of \(200 \mathrm{~N}\) to the lower end. The weight stretches the wire by \(1 \mathrm{~mm}\). Then the elastic energy stored in the wire is [2003] (A) \(0.2 \mathrm{~J}\) (B) \(10 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(0.1 \mathrm{~J}\)

A block of mass \(m\) is slowly bulled up on inclined plane of neight \(h\) and inclination \(\theta\) with he top of a string parallel to the ncline. Which of the following statement is correct for the block when it moves up from the bottom to the top of the incline? A) Work done by the normal reaction force is zero. B) Work done by the string on block is \(m g h\). C) Work done by the gravity is \(m g h .\) D) Work done by the block is \(-m g h / 2\).

A train starts from rest at \(t=0\) along a straight track with a constant acceleration of \(5 \mathrm{~m} / \mathrm{s}^{2}\). A passenger at rest in train observes a particle of mass \(1 \mathrm{~kg}\) on the floor with which it has a co-efficient of friction \(\mu_{s}=\mu_{k}=0.6 .\) At \(t=4 s\), a horizontal force \(F=13 \mathrm{~N}\) is applied on the particle for \(2 \mathrm{~s}\) duration. The passenger observes that the particle is now moving in a perpendicular direction of motion of the train. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) The direction in which force \((F=13 \mathrm{~N})\) is applied is at (A) \(90^{\circ}\) with the direction of motion of the train. (B) \(\cos ^{-1} \frac{5}{13}\) with the direction of motion of the train. (C) \(\cos ^{-1} \frac{5}{12}\) with the direction of motion of the train. (D) \(\cos ^{-1} \frac{12}{13}\) with the direction of motion of the train.

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