Question

A box with an initial speed of `3 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `1/3` and an incline of `( pi )/3`. How far along the ramp will the box go?

Answer 1

Here,as the tendency of the block is to move upwards,hence the frictional force will act along with the component of its weight along the plane to decelerate its motion.

So,net force acting downwards along the plane is `(mg sin ((pi)/3) + mu mg cos ((pi)/3))`

So,net deceleration will be `((g sqrt(3))/2 + 1/3 g (1/2))=10.12 ms^-2`

So,if it moves upward along the plane by `xm` then we can write,

`0^2 =3^2 -2×10.12×x` (using, `v^2=u^2 -2as` and after reaching maximum distance,velocity will become zero)

So, `x=0.45m`

Answer 2

The distance is `=0.44m`

Explanation:

Resolving in the direction up and parallel to the plane as positive `↗^+`

The coefficient of kinetic friction is `mu_k=F_r/N`

Then the net force on the object is

`F=-F_r-Wsintheta`

`=-F_r-mgsintheta`

`=-mu_kN-mgsintheta`

`=mmu_kgcostheta-mgsintheta`

According to Newton's Second Law of Motion

`F=m*a`

Where `a` is the acceleration of the box

So

`ma=-mu_kgcostheta-mgsintheta`

`a=-g(mu_kcostheta+sintheta)`

The coefficient of kinetic friction is `mu_k=1/3`

The acceleration due to gravity is `g=9.8ms^-2`

The incline of the ramp is `theta=1/3pi`

The acceleration is `a=-9.8*(1/3cos(1/3pi)+sin(1/3pi))`

`=-10.12ms^-2`

The negative sign indicates a deceleration

Apply the equation of motion

`v^2=u^2+2as`

The initial velocity is `u=3ms^-1`

The final velocity is `v=0`

The acceleration is `a=-10.12ms^-2`

The distance is `s=(v^2-u^2)/(2a)`

`=(0-9)/(-2*10.12)`

`=0.44m`