(D) According to the law of conservation of mechanical energy,the total mechanical energy at point $A$ is equal to the total mechanical energy at poin

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

(D) According to the law of conservation of mechanical energy,the total mechanical energy at point $A$ is equal to the total mechanical energy at point $B$.$E_A = E_B$$PE_A + KE_A = PE_B + KE_B$Given that the particle starts from rest at $A$,$KE_A = 0$.$mgh_A + 0 = mgh_B + \frac{1}{2}mv^2$$gh_A = gh_B + \frac{1}{2}v^2$$v^2 = 2g(h_A - h_B)$Substituting the given values: $g = 10 \ m/s^2$,$h_A = 10 \ m$,$h_B = 5 \ m$.$v^2 = 2 \times 10 \times (10 - 5)$$v^2 = 20 \times 5 = 100$$v = \sqrt{100} = 10 \ m/s$Therefore,the value of $x$ is $10$.

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

Medium

As shown in the figure,a particle of mass $10 \ kg$ is placed at a point $A$. When the particle is slightly displaced to its right,it starts moving and reaches the point $B$. The speed of the particle at $B$ is $x \ m/s$. (Take $g = 10 \ m/s^2$). The value of $x$ to the nearest integer is:

JEE MAIN 2021 All JEE MAIN

A
$5$
B
$8$
C
$12$
D
$10$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy and Power

Subtopic

Work Energy Theorem and Conservation of Mechanical Energy

Previous Year

JEE MAIN 2021 Paper All JEE MAIN years →

$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

Difficult

View Solution

$A$ mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is

Medium

View Solution

$A$ particle of mass $m \ kg$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to

MediumTS EAMCET 2018

View Solution

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

DifficultKCET 2012

View Solution

$A$ body falls freely from a height $h$ in a gravitational field. Match the following:

Column-$I$ Column-$II$

$(1)$ Potential Energy $=$ Kinetic Energy $(a)$ At height $\frac{2h}{3}$

$(2)$ Potential Energy $= 2 \times$ Kinetic Energy $(b)$ Constant at every point

$(c)$ At height $\frac{h}{2}$

Medium

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

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$

As shown in the figure,a particle of mass $10 \ kg$ is placed at a point $A$. When the particle is slightly displaced to its right,it starts moving and reaches the point $B$. The speed of the particle at $B$ is $x \ m/s$. (Take $g = 10 \ m/s^2$). The value of $x$ to the nearest integer is:

Similar Questions

$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

$A$ mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is

$A$ particle of mass $m \ kg$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

$A$ body falls freely from a height $h$ in a gravitational field. Match the following: Column-$I$ Column-$II$ $(1)$ Potential Energy $=$ Kinetic Energy $(a)$ At height $\frac{2h}{3}$ $(2)$ Potential Energy $= 2 \times$ Kinetic Energy $(b)$ Constant at every point $(c)$ At height $\frac{h}{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ body falls freely from a height $h$ in a gravitational field. Match the following:

Column-$I$ Column-$II$

$(1)$ Potential Energy $=$ Kinetic Energy $(a)$ At height $\frac{2h}{3}$

$(2)$ Potential Energy $= 2 \times$ Kinetic Energy $(b)$ Constant at every point

$(c)$ At height $\frac{h}{2}$