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_k*μ* *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 = V_{1}

Final velocity of the block = V_{2} = 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

V_{1} = 5.785 m/s

From the free body diagram,

N = mg

f =
_{k}N

f =
_{k}mg

ma = f

ma =
_{k}mg

a =
_{k}g

a = (0.503)(9.81)

a = 4.93 m/s^{2}

a = -4.93 m/s^{2} (Negative sign as the block is
decelerating)

Distance the block slides horizontally before it stops = d

V_{2}^{2} = V_{1}^{2} + 2ad

0^{2} = 5.785^{2} + 2(-4.93)d

d = 3.39 m

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

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