Question

A variable force $F$ is applied on a block such that $F\propto t$, at an angle $\theta$ with the horizontal as shown in figure. Given $({\mu }_s>{\mu }_k)$. Find the correct graph between frictional force and time. (where $t=$ time)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given, $F\propto t$
Force $F$ is increasing with time $t$.
From FBD:


$N=mg-F\sin \theta$
$f=F\cos \theta$
Block is initially at rest,
$\therefore$ At $t=0,f=0$
Hence, graph will start from origin and increase linearly.
i.e $f\propto t$

Maximum static frictional force that can act,
$(f_s{)}_{max}={\mu }_{s}N$
$(f_s{)}_{max}={\mu }_s(mg-F\sin \theta )$
At time $t$ when maximum static friction $(f_s{)}_{max}$ will be equal to $F\cos \theta$, then block will just start moving.

When block starts moving, kinetic friction will act
$f_k={\mu }_{k}N{f}_k={\mu }_k(mg-F\sin \theta )---\left(1\right)$
As, given ${\mu }_k<{\mu }_s$. Hence $(f_k<f_s)$

Here, $F$ is increasing linearly with time.
Hence, $f_k$ will decrease linearly with time and will eventually become zero according to equation $\left(1\right)$.
So, graph will decrease linearly, after attending the maximum value.
Hence option (a) is correct.