(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.For the first case,

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 body.For the first case,the body comes to rest from velocity $u$ after covering distance $s_{1}$:$F s_{1} = \frac{1}{2} m u^{2} \quad (i)$For the second case,the body comes to rest from velocity $2u$ after covering distance $s_{2}$:$F s_{2} = \frac{1}{2} m (2u)^{2} = \frac{1}{2} m (4u^{2}) = 4 \left( \frac{1}{2} m u^{2} \right) \quad (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{F s_{2}}{F s_{1}} = \frac{4 (\frac{1}{2} m u^{2})}{\frac{1}{2} m u^{2}}$$\frac{s_{2}}{s_{1}} = 4$$s_{2} = 4s_{1}$

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

Difficult

$A$ 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,

KCET 2012 All KCET

A
$s_{2} = 4s_{1}$
B
$s_{2} = 2s_{1}$
C
$s_{2} = \frac{s_{1}}{2}$
D
$s_{2} = s_{1}$

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

KCET 2012 Paper All KCET years →

$A$ body dropped from height $H$ reaches the ground with a speed of $1.1 \sqrt{gH}$. Calculate the work done by air friction in terms of $mgH$.

Difficult

View Solution

$A$ particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \ m/s^2$).

DifficultJEE MAIN 2024

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

$A$ car travelling at $80 \ km/h$ can be stopped at a distance of $60 \ m$ by applying brakes. If the same car travels at $160 \ km/h$ and the same braking force is applied,the stopping distance is: (in $m$)

EasyAP EAMCET 2024

View Solution

$A$ particle of mass $m$ is moving along a circle of radius $R$ such that its tangential acceleration $a_t$ varies with distance covered $x$ as $a_t = \alpha x^2$,where $\alpha$ is a constant. The kinetic energy $K$ of the particle varies with the distance as $K = \beta x^c$,where $\beta$ and $c$ are constants. The values of $\beta$ and $c$ are:

MediumTS EAMCET 2019

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

$A$ 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,

Similar Questions

$A$ body dropped from height $H$ reaches the ground with a speed of $1.1 \sqrt{gH}$. Calculate the work done by air friction in terms of $mgH$.

$A$ particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \ m/s^2$).

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 $g = 10 \text{ m/s}^2$)

$A$ car travelling at $80 \ km/h$ can be stopped at a distance of $60 \ m$ by applying brakes. If the same car travels at $160 \ km/h$ and the same braking force is applied,the stopping distance is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module