Question

A block of mass, 25.7 kg, starts at rest at the top of a frictionless ramp that makes an angle of 45.6 ^\circ ​∘ ​​ below the horizontal. After it slides without friction down the entire length of the ramp, it begins to slide horizontally along a rough concrete surface with a coefficient of kinetic friction of \mu_kμ ​k ​​ = 0.545 until it slows to a complete stop. The initial height of the block at the top of the ramp is 1.79 meters. How far does the block slide horizontally along the concrete before it stops? [Note : this question may contain more information than is necessary to solve the problem.]

Answer 1

Before the block stops moving a distance of 3.28 m

Step-by-step explanation:

Given data:

m = 25.7 kg

μk = 0.545

h = 1.79 m

The kinetic energy at point B is equal:

`E_(kB) =mgh=25.7*9.8*1.79=450.83J`

The velocity at point B is equal:

`v=√(2gh) =√(2*9.8*1.79) =5.92m/s`

The friction force is equal:

`f_(k) =\mu _(k) mg=ma\\`

Clearing a:

`a=\mu _(k) g=0.545*9.8=5.341m/s^(2)`

The horizontal distance is:

`L=(v^(2) )/(2a) =(5.92^(2) )/(2*5.341) =3.28m`