Question

Force acting on a block versus time graph is as shown in figure. Choose the correct options. (Take m=2kg and g = 10 $m{s}^{-2}$)

• A. At t = 2 s, force of friction is 2 N.
• B. At t = 8 s, force of friction is 8 N.
• C. At t = 10 s, acceleration of block is $2m{s}^{-2}$
• D. At t = 12s, velocity of block is $8m{s}^{-1}$

Answer 1

Answer: (A), (C), (D)

At t = 2 s, force of friction is 2 N.
At t = 10 s, acceleration of block is $2m{s}^{-2}$
At t = 12s, velocity of block is $8m{s}^{-1}$

Statc friction force

$f_s\le \mu N$

N=2X10 = 20

$f_s\le 0.3\times 20f_s\le 6N$

As shown in graph, thus static friction is equal to force applied upto $6N$ at $t=2s$ force applied $=2N$ = Friction at time $t=6s$ force $=6N$ and friction is also $6N$

Now motion of particle will start

at $t=8s$ force $=8N$ it is greater than static friction, then friction is limiting $6N$

Acceleration of particle is $a=\frac{8-6}{2}=1m/s^2$

At time $t=10s$

Force applied is $10N$

Acceleration

$a=\frac{10-6}{2}=2m/s^2$

Velocity at time $t=10s$

$a=\frac{t-6}{2}\frac{dv}{dt}=\frac{t-6}{2}2dv=(t-6)dt$

By integrating both sides

$v=\frac{t^2}{4}-3t+c$ at $t=6$$v=0$ put these values we get $c=9$

Velocity as a function of time

$v=\frac{t^2}{4}-3t+9$

At time $t=10$ we get velocity $4m/s$

At time $t=12$ acceleration is constant $a=2$

$v=u+atatt=10v=4v=4+2\times (12-10)v=8m/s$

All options are correct

A,C,D are correct