Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Consider a block kept on an inclined plane (inclined at ) as shown in the figure. If the force

Consider a block kept on an inclined plane (inclined at 45^{\circ}) as shown in the figure. If the force
required to just push it up the incline is 2 times the force required to just prevent it from sliding down,
the coefficient of friction between the block and inclined plane (\mu) is equal to :

Option: 1

0.25


Option: 2

0.50


Option: 3

0.60


Option: 4

0.33


Answers (1)

 
\begin{aligned} & \mathrm{F}_1=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta \\ & \mathrm{F}_1=\mathrm{mg} \sin 45+\mu \mathrm{mg} \cos 45 \\ & \mathrm{~F}_2=\mathrm{mg} \sin 45-\mu \mathrm{mg} \cos 45 \end{aligned}
\begin{aligned} & \mathrm{F}_1=2 \mathrm{~F}_2 \\ & \mathrm{mg}\left(\frac{1}{\sqrt{2}}+\frac{\mu}{\sqrt{2}}\right)=2 \mathrm{mg}\left(\frac{1}{\sqrt{2}}-\frac{\mu}{\sqrt{2}}\right) \\ & 1+\mu=2-2 \mu \\ & 3 \mu=1 \\ & \mu=\frac{1}{3}=0.33 \end{aligned}

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question