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.]

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lcoley8 lcoley8 · Answer 1 · 2020-04-20T05:18:24+02:00

Answer:

Before the block stops moving a distance of 3.28 m

Explanation:

Given data:

m = 25.7 kg

μk = 0.545

h = 1.79 m

The kinetic energy at point B is equal:

[tex]E_{kB} =mgh=25.7*9.8*1.79=450.83J[/tex]

The velocity at point B is equal:

[tex]v=\sqrt{2gh} =\sqrt{2*9.8*1.79} =5.92m/s[/tex]

The friction force is equal:

[tex]f_{k} =\mu _{k} mg=ma\\[/tex]

Clearing a:

[tex]a=\mu _{k} g=0.545*9.8=5.341m/s^{2}[/tex]

The horizontal distance is:

[tex]L=\frac{v^{2} }{2a} =\frac{5.92^{2} }{2*5.341} =3.28m[/tex]