Question

A block of mass 19.6 kg starts at rest at the top of a frictionless ramp that makes an angle of 36.2 ^\circ ∘ below the horizontal. After it slides without friction down the entire 2.89 m length of the ramp, it begins to slide horizontally along a rough concrete surface with a coefficient of kinetic friction of \mu_kμ k = 0.503 until it slows to a complete stop. 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

Mass of block = m = 19.6 kg

Angle of ramp = = 36.2^o below the horizontal

Length of ramp = L = 2.89 m

Height of ramp = h

Coefficient of kinetic friction on rough horizontal floor = _k = 0.503

Velocity of block at the end of the ramp = V₁

Final velocity of the block = V₂ = 0 m/s (It slows to a complete stop)

From the figure,

Sin = h/L

Sin(36.2) = h/(2.89)

h = 1.706 m

The potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom of the ramp

V₁ = 5.785 m/s

From the free body diagram,

N = mg

f = _kN

f = _kmg

ma = f

ma = _kmg

a = _kg

a = (0.503)(9.81)

a = 4.93 m/s²

a = -4.93 m/s² (Negative sign as the block is decelerating)

Distance the block slides horizontally before it stops = d

V₂² = V₁² + 2ad

0² = 5.785² + 2(-4.93)d

d = 3.39 m

The block slides 3.39m horizontally along the concrete before it stops.