(C) The work done by a force $F$ on a particle is given by $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt$.Since the force is always

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The work done by a force $F$ on a particle is given by $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt$.Since the force is always perpendicular to the velocity,$\vec{F} \cdot \vec{v} = 0$.Therefore,the work done $W = 0$.According to the work-energy theorem,the change in kinetic energy $\Delta K = W$.Since $W = 0$,the change in kinetic energy $\Delta K = 0$,which implies that the kinetic energy of the particle remains constant.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Easy

$A$ particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

AIEEE 2004 All AIEEE

A
Its velocity is constant
B
Its acceleration is constant
C
Its kinetic energy is constant
D
It moves in a straight line

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy, Power and Collision

Subtopic

Work Energy Theorem and Conservation of Mechanical Energy

Previous Year

AIEEE 2004 Paper All AIEEE years →

$A$ particle of mass $m$ moves with a velocity $v = a\sqrt{x}$,where $a$ is a constant and $x$ is the displacement. Find the total work done by all forces during the displacement from $x = 0$ to $x = d$.

Medium

View Solution

The velocity at which the $6 \text{ kg}$ mass (shown in the figure) strikes the ground when it is released from a height of $6 \text{ m}$ above the ground is . . . . . . $\text{m/s}$. Assume the pulley is massless and the string is light and inextensible. (Take $g = 10 \text{ m/s}^2$)

DifficultJEE MAIN 2026

View Solution

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively:

EasyAP EAMCET 2018

View Solution

What is the velocity of the bob when it reaches point $B$ in $m/s$?

Medium

View Solution

From a height of $h$ above the ground, a ball is projected up at an angle $30^{\circ}$ with the horizontal. If the ball strikes the ground with a speed of $1.25$ times its initial speed of $40 \,ms^{-1}$, the value of $h$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

MediumTS EAMCET 2024

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

Similar Questions

$A$ particle of mass $m$ moves with a velocity $v = a\sqrt{x}$,where $a$ is a constant and $x$ is the displacement. Find the total work done by all forces during the displacement from $x = 0$ to $x = d$.

The velocity at which the $6 \text{ kg}$ mass (shown in the figure) strikes the ground when it is released from a height of $6 \text{ m}$ above the ground is . . . . . . $\text{m/s}$. Assume the pulley is massless and the string is light and inextensible. (Take $g = 10 \text{ m/s}^2$)

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively:

What is the velocity of the bob when it reaches point $B$ in $m/s$?

From a height of $h$ above the ground, a ball is projected up at an angle $30^{\circ}$ with the horizontal. If the ball strikes the ground with a speed of $1.25$ times its initial speed of $40 \,ms^{-1}$, the value of $h$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module