(A) The retardation $a$ of the block is given by $a = \frac{dv}{dt} = -\mu g$.Integrating this expression,we get $\Delta v = v_f - v_i = -g \int_{0}^{

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The retardation $a$ of the block is given by $a = \frac{dv}{dt} = -\mu g$.Integrating this expression,we get $\Delta v = v_f - v_i = -g \int_{0}^{4} \mu \,dt$.The term $\int_{0}^{4} \mu \,dt$ represents the area under the $\mu-t$ curve from $t = 0$ to $t = 4 \,s$.The area consists of a rectangle from $t = 0$ to $2 \,s$ and a trapezoid from $t = 2$ to $4 \,s$.Area $= (2 \times 0.5) + \frac{1}{2} \times (0.5 + 0.3) \times (4 - 2) = 1 + 0.8 = 1.8$.Now,substituting the values into the velocity equation:$v_f - 20 = -10 \times (1.8) = -18$.$v_f = 20 - 18 = 2 \,m/s$.

Class 11 Physics Laws of Motion Kinetic Friction and Motion on Rough Horizontal Surface

Difficult

$A$ block is projected with speed $20 \,m/s$ on a rough horizontal surface. The coefficient of friction $(\mu)$ between the surfaces varies with time $(t)$ as shown in the figure. The speed of the body at the end of $4 \,s$ will be ............ $m/s$ $(g = 10 \,m/s^2)$.

A
$2$
B
$5$
C
$7.2$
D
$9.5$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Laws of Motion

Subtopic

Kinetic Friction and Motion on Rough Horizontal Surface

$A$ block of mass $10 \, kg$ moving with an acceleration of $2 \, m/s^2$ on a horizontal rough surface is shown in the figure. The value of the coefficient of kinetic friction is ...........

Medium

View Solution

$A$ body of mass $2$ kg is moving on the ground and comes to rest after some time. The coefficient of kinetic friction between the body and the ground is $0.2$. The retardation in the body is ...... $m/s^2$.

Easy

View Solution

$A$ block of mass $4\,kg$ is placed on a rough horizontal plane. $A$ time-dependent force $F = kt^2$ acts on the block,where $k = 2\,N/s^2$. The coefficient of friction is $\mu = 0.8$. The force of friction between the block and the plane at $t = 2\,s$ is ....... $N$.

Medium

View Solution

$A$ body of mass $2 \, kg$ moving on a horizontal surface with an initial velocity of $4 \, m/s$ comes to rest after $2 \, s$. If one wants to keep this body moving on the same surface with a constant velocity of $4 \, m/s$,the force required is ........ $N$.

Medium

View Solution

$A$ body is pulled along a rough horizontal surface with a velocity $6\,m/s$. If the body comes to rest after travelling $9\,m$,then the coefficient of sliding friction is- (Take $g = 10\,m/s^2$)

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ block is projected with speed $20 \,m/s$ on a rough horizontal surface. The coefficient of friction $(\mu)$ between the surfaces varies with time $(t)$ as shown in the figure. The speed of the body at the end of $4 \,s$ will be ............ $m/s$ $(g = 10 \,m/s^2)$.

Similar Questions

$A$ block of mass $10 \, kg$ moving with an acceleration of $2 \, m/s^2$ on a horizontal rough surface is shown in the figure. The value of the coefficient of kinetic friction is ...........

$A$ body of mass $2$ kg is moving on the ground and comes to rest after some time. The coefficient of kinetic friction between the body and the ground is $0.2$. The retardation in the body is ...... $m/s^2$.

$A$ block of mass $4\,kg$ is placed on a rough horizontal plane. $A$ time-dependent force $F = kt^2$ acts on the block,where $k = 2\,N/s^2$. The coefficient of friction is $\mu = 0.8$. The force of friction between the block and the plane at $t = 2\,s$ is ....... $N$.

$A$ body of mass $2 \, kg$ moving on a horizontal surface with an initial velocity of $4 \, m/s$ comes to rest after $2 \, s$. If one wants to keep this body moving on the same surface with a constant velocity of $4 \, m/s$,the force required is ........ $N$.

$A$ body is pulled along a rough horizontal surface with a velocity $6\,m/s$. If the body comes to rest after travelling $9\,m$,then the coefficient of sliding friction is- (Take $g = 10\,m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module