Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A 2 kg mass starts from rest on an inclined smooth surface with inclination $30^{o}$ and length 2 m. How much will it travel (in meters) before coming to rest on a surface with coefficient of friction 0.25?

Option: 1
4

Option: 2
6

Option: 3
8

Option: 4
2

Answers (1)

best_answer

As we learned

On the horizontal road

A block of mass m is moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance S.

Distance traveled before coming to rest (S):

$F=ma = \mu R$

$ma = \mu mg$

$a = \mu g$

$V^{2}=u^{2}-2as$

$S = \frac{u^{2}}{2\mu g}=\frac{P^{2}}{2\mu m^{2}g}$

a = acceleration

$\mu=$ coefficient of friction

S = distance traveled

g = gravity

u=initial velocity

V = finally velocity

P=initial mometum=mu

On the inclined road :

$a=g[sin\theta+\mu cos\theta]$

$V^{2}=u^{2}-2aS$

$0=u^{2}-2g[sin\theta+\mu cos \theta]S$

$S=\frac{u^{2}}{2g(sin\theta+\mu cos\theta)}$

S= distance traveled

$\mu=$ coefficient of friction

V = Final velocity

u = Initial velocity

By using this concept,

So First on incline plane

$v^{2}=u^{2}+2as=0+2\times g\; \sin 30\times 2$

Let it travel distance 'S' before coming to rest on a horizontal surface

$S=\frac{v^{2}}{2\mu g}=\frac{20}{2\times 0.25\times 10}$

$= 4m$

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

Andhra launches free JEE, NEET coaching to over 1 lakh govt school students

anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions

anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel

Latest Question