Get 2 FREE Instant-Explanations on Filo with code FILOAPP

Home

Physics Part-I

Laws of motion

A constant retarding force of 50 N is applied to a body of mas

Physics Part-I

Class 11

NCERT

Chapter 1: Physical World

16 questions

Chapter 2: Units and Measurements

51 questions

Chapter 3: Motion in Straight Line

50 questions

Chapter 4: Motion in a Plane

43 questions

Chapter 5: Laws of motion

52 questions

One end of a string of length

l

is connected to a particle of mass

m

and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed

v

the net force on the particle ( directed towards the centre ) is :

A constant retarding force of

50

N

is applied to a body of mass

20

k g

moving initially with a speed of

15

m s^{- 1}

. How long does the body take to stop?

A constant force acting on a body of mass

3.0 k g

changes its speed from

2.0 m

s^{- 1}

3.5 m

s^{- 1}

25 s

. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

View all question

Chapter 6: Work, Energy and Power

41 questions

Chapter 7: System of Particles and Rotational Motion

45 questions

Chapter 8: Gravitation

33 questions

Question

Easy

Solving time: 1 mins

A constant retarding force of $50$ $N$ is applied to a body of mass $20$ $k g$ moving initially with a speed of $15$ $m s^{- 1}$ . How long does the body take to stop?

Views: 6,082 students

Updated on: Feb 21, 2024

Book:

Physics Part-I (NCERT)

Found 6 tutors discussing this question

Michael Discussed

A constant retarding force of

50

N

is applied to a body of mass

20

k g

moving initially with a speed of

15

m s^{- 1}

. How long does the body take to stop?

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Retarding force,

F = 50 N

Mass of the body,

m = 20 k g

Initial velocity of the body,

u = 15 m / s

Final velocity of the body,

v = 0

Using Newtons second law of motion, the acceleration (

a

) produced is given by:

F = ma

50 = 20

\times a

a = - 50/20 = - 2.5 m / s

^{2}

Using the first equation of motion, the time (

t

) taken by the body to come at rest

v = u + a t

t = - u / a = - 15/ - 2.5 = 6 s

Was this solution helpful?

Report

Filo tutor solutions (28)

Learn from their 1-to-1 discussion with Filo tutors.

4 mins

Uploaded on: 8/27/2023

Ask your question, on a video call with tutor

Connect instantly with this tutor

Connect now

Taught by

Ubaid

Total classes on Filo by this tutor - 12,615

Teaches : Physics, Mathematics, Chemistry

Connect instantly with this tutor

Connect now

Notes from this class (1 pages)

Download

Was this solution helpful?

Report

3 mins

Uploaded on: 12/6/2023

Ask your question, on a video call with tutor

Connect instantly with this tutor

Connect now

Taught by

Piyush

Total classes on Filo by this tutor - 3,407

Teaches : Physics, Mathematics

Connect instantly with this tutor

Connect now

Notes from this class (3 pages)

Download

Was this solution helpful?

Report

3 mins

Uploaded on: 7/11/2023

Ask your question, on a video call with tutor

Connect instantly with this tutor

Connect now

Taught by

SHAHZER HANEEF

Total classes on Filo by this tutor - 9,071

Teaches : Science, Mathematics, English

Connect instantly with this tutor

Connect now

Notes from this class (2 pages)

Download

Was this solution helpful?

122

Report

See more 1-on-1 discussions (25)

One destination for complete JEE/NEET preparation

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Practice more questions from Physics Part-I (NCERT)

One end of a string of length

l

is connected to a particle of mass

m

and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed

v

the net force on the particle ( directed towards the centre ) is :

A constant retarding force of

50

N

is applied to a body of mass

20

k g

moving initially with a speed of

15

m s^{- 1}

. How long does the body take to stop?

A constant force acting on a body of mass

3.0 k g

changes its speed from

2.0 m

s^{- 1}

3.5 m

s^{- 1}

25 s

. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

View all

Stuck on the question or explanation?

Connect with our Physics tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	A constant retarding force of $50$ $N$ is applied to a body of mass $20$ $k g$ moving initially with a speed of $15$ $m s^{- 1}$ . How long does the body take to stop?
Updated On	Feb 21, 2024
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 28
Upvotes	3049
Avg. Video Duration	4 min

A constant retarding force of $50$ $N$ is applied to a body of mass $20$ $k g$ moving initially with a speed of $15$ $m s^{- 1}$ . How long does the body take to stop?

Text solutionVerified

Filo tutor solutions (28)

Practice more questions from Physics Part-I (NCERT)

Practice questions from Physics Part-I (NCERT)

Practice more questions from Laws of motion

Practice questions on similar concepts asked by Filo students

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms−1. How long does the body take to stop?

Text solutionVerified

Filo tutor solutions (28)

Practice more questions from Physics Part-I (NCERT)

Practice questions from Physics Part-I (NCERT)

Practice more questions from Laws of motion

Practice questions on similar concepts asked by Filo students

Related books for questions on Laws of motion

A constant retarding force of $50$ $N$ is applied to a body of mass $20$ $k g$ moving initially with a speed of $15$ $m s^{- 1}$ . How long does the body take to stop?