New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Laws of Motion question

2023 · 1 Feb · Shift 1 · Q57

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Physics
Laws of Motion
2023 · 1 Feb · Shift 1 · Q57

Laws of Motion question

2023 · 1 Feb · Shift 1 · Q57

JEE MainPhysicsLaws of MotionMCQ+4 / −1

A block of mass

5 \mathrm{~kg}

is placed at rest on a table of rough surface. Now, if a force of

30 \mathrm{~N}

is applied in the direction parallel to surface of the table, the block slides through a distance of

50 \mathrm{~m}

in an interval of time

10 \mathrm{~s}

. Coefficient of kinetic friction is (given,

g=10 \mathrm{~ms}^{-2}

):

A
0.25
B
0.75
C
0.60
D
0.50

View written solutionFree

Correct answer: D

Given data

Mass of block: $m = 5\,\text{kg}$
Applied force: $F = 30\,\text{N}$
Distance moved: $s = 50\,\text{m}$
Time taken: $t = 10\,\text{s}$
Initial velocity: $u = 0$ (block starts from rest)
$g = 10\,\text{m s}^{-2}$

Find acceleration using kinematics

Since the block starts from rest,

$s = ut + \frac{1}{2}at^2$

Substitute $u=0$ :

$50 = \frac{1}{2}a(10)^2$

$50 = 50a$

$a = 1\,\text{m s}^{-2}$

Apply Newton's second law

The horizontal forces on the block are:

Applied force = $30\,\text{N}$
Kinetic friction = $f_k$

So,

$F - f_k = ma$

$30 - f_k = 5 \times 1$

$30 - f_k = 5$

$f_k = 25\,\text{N}$

Use formula for kinetic friction

$f_k = \mu_k N$

Since the force is horizontal, normal reaction is

$N = mg = 5 \times 10 = 50\,\text{N}$

Thus,

$25 = \mu_k \times 50$

$\mu_k = \frac{25}{50} = 0.5$

Check options

A: $0.25$ ❌
B: $0.75$ ❌
C: $0.60$ ❌
D: $0.50$ ✅

Therefore, the correct answer is Option D.

More from Laws of Motion

As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle $3 0^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of F : [Given $g = 10 ms^{- 2}$ … Includes diagram2023 · MCQ
Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure. Includes diagram2023 · MCQ
At any instant the velocity of a particle of mass $(2 t \hat{i} + 3 t^{2} \hat{j}) ms^{- 1}$ is $(2 t \hat{i} + 3 t^{2} \hat{j}) ms^{- 1}$ . If the force acting on the particle at $t = 1 s$ is $(\hat{i} + x \hat{j}) N$ . Then the…2023 · MCQ
A body of mass $x$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 x m / s$ the force acting on the body is:-2023 · MCQ
Three forces $F_{1} = 10 N, F_{2} = 8 N, F_{3} = 6 N$ are acting on a particle of mass $F_{2}$ . The forces $F_{2}$ and $F_{3}$ are applied perpendicularly so that particle…2023 · MCQ
The position vector of a particle related to time $t$ is given by $r = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m$ The direction of net force experienced by the particle is :2023 · MCQ
As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s $^{2}$ ) Includes diagram2023 · MCQ
Given below are two statements : Statement I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II : Force exerted by the floor of an elevator on the foot of a person…2023 · MCQ