(A) According to the work-energy theorem, the work done by the retarding force is equal to the change in kinetic energy of the bus.$W = \Delta K$$F \c

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

(A) According to the work-energy theorem, the work done by the retarding force is equal to the change in kinetic energy of the bus.$W = \Delta K$$F \cdot s = \frac{1}{2} m v^2$Since the retarding force $F$ and the initial velocity $v$ are constant, we have:$s = \frac{m v^2}{2 F} \implies s \propto m$Let the initial mass be $m_1 = m$ and the initial stopping distance be $s_1 = x$.The new mass after increasing by $25\%$ is $m_2 = m + 0.25m = 1.25m$.Using the proportionality $s \propto m$:$\frac{s_2}{s_1} = \frac{m_2}{m_1}$$s_2 = s_1 \cdot \frac{1.25m}{m}$$s_2 = 1.25 x$.

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

Difficult

$A$ bus moving on a level road with a velocity $v$ can be stopped at a distance of $x$ by the application of a retarding force $F$. The load on the bus is increased by $25\%$ by boarding the passengers. Now, if the bus is moving with the same speed and if the same retarding force is applied, the distance travelled by the bus before it stops is

TS EAMCET 2014 All TS EAMCET

AP EAMCET 2014 All AP EAMCET

A
$1.25 x$
B
$x$
C
$5 x$
D
$2.5 x$

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

TS EAMCET 2014 Paper All TS EAMCET years →

Previous Year

AP EAMCET 2014 Paper All AP EAMCET years →

$A$ body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 \sqrt{gh}$. The value of work done by the air friction is $.....$

DifficultJEE MAIN 2021

View Solution

$A$ pendulum bob has a speed of $3 \, m/s$ at its lowest position. The pendulum is $50 \, cm$ long. The speed of the bob, when the string makes an angle of $60^{\circ}$ to the vertical, will be $....... \, m/s$ $(g = 10 \, m/s^2)$.

MediumJEE MAIN 2021

View Solution

Three particles $A, B$, and $C$ are thrown from the top of a tower with the same speed $u$. Particle $A$ is thrown upwards, $B$ is thrown downwards, and $C$ is thrown horizontally. If they hit the ground with speeds $V_A, V_B$, and $V_C$ respectively, then:

Medium

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

The figure shows the vertical section of a frictionless surface. $A$ block of mass $2\, kg$ is released from position $A$. Its kinetic energy $(KE)$ as it reaches position $C$ is .............. $J$.

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

Similar Questions

$A$ body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 \sqrt{gh}$. The value of work done by the air friction is $.....$

$A$ pendulum bob has a speed of $3 \, m/s$ at its lowest position. The pendulum is $50 \, cm$ long. The speed of the bob, when the string makes an angle of $60^{\circ}$ to the vertical, will be $....... \, m/s$ $(g = 10 \, m/s^2)$.

Three particles $A, B$, and $C$ are thrown from the top of a tower with the same speed $u$. Particle $A$ is thrown upwards, $B$ is thrown downwards, and $C$ is thrown horizontally. If they hit the ground with speeds $V_A, V_B$, and $V_C$ respectively, then:

$A$ 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,

The figure shows the vertical section of a frictionless surface. $A$ block of mass $2\, kg$ is released from position $A$. Its kinetic energy $(KE)$ as it reaches position $C$ is .............. $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module