Question

A box of mass m is initially at rest on a horizontal surface. A constant horizontal force of mg/2 is applied to the box directed to the right. The coefficient of friction of the surface changes with the distance pushed as μ=μ0x where x is the distance from the initial location. For what distance is the box pushed until it comes to rest again ?

Answer 1

initial velocity of box , u = 0
Let final velocity of box is v
it is given that constant horizontal force , f = mg/2
coefficient of friction varies with x as $\mu=\mu_0x$

when body moves right direction due to constant horizontal force, friction acts just opposite direction. e.g., left direction.
so, net force acts body body at any instant is given by $F_{net}=f-f_s$

we know, from Newton's 2nd law,
$F_{net}=ma=mv\frac{dv}{dx}$

so, $mv\frac{dv}{dx}=\frac{mg}{2}-\mu N$
here N is normal reaction act on box,
so, N = mg
$mv\frac{dv}{dx}=\frac{mg}{2}-\mu_0xmg$

$2\int\limits^v_u{v}\,dv=\int\limits^x_0{(g-2\mu_0gx)}\,dx$

[v² - u² ] = gx - $\mu_0$gx²

u = 0, so, v² = gx - $\mu_0$gx²
body comes to rest when , v = 0
so, x = $\frac{1}{\mu_0}$

Answer 2

Answer:

Explanation:

initial velocity of box , u = 0

Let final velocity of box is v

it is given that constant horizontal force , f = mg/2

coefficient of friction varies with x as  

when body moves right direction due to constant horizontal force, friction acts just opposite direction. e.g., left direction.

so, net force acts body body at any instant is given by  

we know, from Newton's 2nd law,

so,  

here N is normal reaction act on box,

so, N = mg

[v² - u² ] = gx - gx²

u = 0, so, v² = gx - gx²

body comes to rest when , v = 0

so, x =