Physics

Friction and Inclined Planes

63 Questions

Friction and inclined plane problems focus on calculating forces, coefficients of static and kinetic friction, and motion on rough surfaces. Questions cover blocks on horizontal tables, angled planes, and simple harmonic motion systems. These physics concepts are strictly evaluated in engineering entrance and civil services tests.

Static friction coefficientKinetic friction forceInclined plane blocksDamped vibrations systemHorizontal table motion

Friction and Inclined Planes Questions

Multiple choice
  1. Only a

  2. Only a and b

  3. Only b

  4. Only b and c

  5. Only c

Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

This is the correct option, as this statement is false. Force of friction is greater, if very rough surfaces are involved.

Multiple choice
  1. P and Q only

  2. P and S only

  3. P, Q and R only

  4. Q and S only

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice
  1. 490.5 and 0.5

  2. 981 and 0.5

  3. 1000.5 and 0.15

  4. 1000.5 and 0.25

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Correct option is (1).

Multiple choice friction laws of motion physics

A block released from rest from the top of a smooth inclined plane of angle $\theta _1$ reaches the bottom in time $t _1$. The same block released from rest from the top of another smooth inclined plane of angle $\theta _2$, reaches the bottom in time $t _2$. If the two inclined planes have the same height, the relation between $t _1$ and $t _2$ is 

  1. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)^{1/2}$
  2. $\cfrac{t _2}{t _1} = 1$
  3. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)$
  4. $\cfrac{t _2}{t _1} = \left(\cfrac{sin^2 \theta _1}{sin^2 \theta _2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
From the figure
we see that
down the incline acceleration us $a=g\sin \theta$
distance $s=\dfrac {h}{\sin \theta _1}$

using $s=ut+\dfrac {1}{2} at^2 \ ;\ u=0$ (initially rest)

gives $t=\sqrt {\dfrac {2s}{a}}$

For $\theta _1 \quad t _1=\sqrt {\dfrac {2\ h}{\sin \theta _1 \times g\sin \theta _1}}=\dfrac {1}{\sin \theta _1} \sqrt {\dfrac {2\ h}{g}}$

so for $O _2, \ t _2=\dfrac {1}{\sin \theta _2}\sqrt {\dfrac {2\ h}{g}}$

$\Rightarrow \ \dfrac {t _2}{t _1}=\dfrac {\sin \theta _1}{\sin \theta _2}$