A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction betw

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively [2015]

1 0.5 and 0.6

2 0.4 and 0.3

3 0.6 and 0.6

4 0.6 and 0.5

Ans.
(4)

Let $μ_{s}$ and $μ_{k}$ be the coefficients of static and kinetic friction between the box and the plank respectively.

When the angle of inclination $θ$ reaches $30 °$ , the block just slides,

$∴ μ_{s} = \tan θ = \tan 30 ° = \frac{1}{\sqrt{3}} = 0.6$

If $a$ is the acceleration produced in the block, then

$m a = m g \sin θ - f_{k}$   (where $f_{k}$ is force of kinetic friction)

   $= m g \sin θ - μ_{k} N$    $(as f_{k} = μ_{k} N)$

   $= m g \sin θ - μ_{k} m g \cos θ$ $(as N = mgcos θ)$

$a = g (\sin θ - μ_{k} \cos θ)$

As    $g = 10 {ms}^{- 2}$ and $θ = 30 °$

$∴$ $a = (10 {ms}^{- 2}) (\sin 30 ° - μ_{k} \cos 30 °)$ ...(i)

If $s$ is the distance travelled by the block in time $t$ , then

$s = \frac{1}{2} a t^{2} (as u = 0) or a = \frac{2 s}{t^{2}}$

But $s = 4.0 m$ and $t = 4.0 s$ (given)

$∴$ $a = \frac{2 (4.0 m)}{(4.0 {s)}^{^{2}}} = \frac{1}{2} {ms}^{- 2}$

Substituting this value of $a$ in equation (i), we get

$\frac{1}{2} {ms}^{- 2} = (10 {ms}^{- 2}) (\frac{1}{2} - μ_{k} \frac{\sqrt{3}}{2})$

$\frac{1}{10} = 1 - \sqrt{3} μ_{k} or \sqrt{3} μ_{k} = 1 - \frac{1}{10} = \frac{9}{10} = 0.9$ or $μ_{k} = \frac{0.9}{\sqrt{3}} = 0.5$

Want Us to Email you About Special Offers & Updates?