2. A 5.00-kg rock piece is released from rest at the top of a 2.50-m-long ramp, reaching a speed of 1.75 m/s at the bottom. The ramp is inclined at an angle of 30°. Calculate the coefficient of kinetic friction between the rock and the ramp. *Do not forget to draw the FBD* Kinematic VxVox + axt 1 x = xo + vox² + ₂a v² = x + 2a, (x-xo) 1 x-xo-(x+₂) = 2 Equations Vy = Voy+at y = yo voyt + 1 29t v=vy + 2a,(y-yo) y - y₁ - 1²/ (1₂+1₂)+ yo = (voy t GEN (Newto Newton's Laws of Motion Σ F = 0 (Newton's 1st Law) 2 F = ma (Newton's 2nd Law) fsmax = Msn fx = Men 2.50m 30° 9 = 9.8 m/s² weight = mass x g

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2. A 5.00-kg rock piece is released from rest at
the top of a 2.50-m-long ramp, reaching a speed
of 1.75 m/s at the bottom. The ramp is inclined at
an angle of 30°. Calculate the coefficient of
kinetic friction between the rock and the ramp.
*Do not forget to draw the FBD*
Kinematic
VxVox + axt
1
x=x0+ x² + x²
voxt
v² = x + 2a, (x-xo)
1
x-xo=(vox + v)t
Equations
Vy = voy + ant
1
y
= 3 + Payt + ayt²
=
v=vy + 2a, (y - Yo)
10 = 1/2 (voy + v₂ ) t
y - yo =
139
(Heat
Newton's Laws of Motion
ΣF = 0
(Newton's 1st Law)
2 F = ma
(Newton's 2nd Law)
fsmax = Msn fk = Men
2.50 m
30°
g = 9.8 m/s²
weight = mass x g
Transcribed Image Text:2. A 5.00-kg rock piece is released from rest at the top of a 2.50-m-long ramp, reaching a speed of 1.75 m/s at the bottom. The ramp is inclined at an angle of 30°. Calculate the coefficient of kinetic friction between the rock and the ramp. *Do not forget to draw the FBD* Kinematic VxVox + axt 1 x=x0+ x² + x² voxt v² = x + 2a, (x-xo) 1 x-xo=(vox + v)t Equations Vy = voy + ant 1 y = 3 + Payt + ayt² = v=vy + 2a, (y - Yo) 10 = 1/2 (voy + v₂ ) t y - yo = 139 (Heat Newton's Laws of Motion ΣF = 0 (Newton's 1st Law) 2 F = ma (Newton's 2nd Law) fsmax = Msn fk = Men 2.50 m 30° g = 9.8 m/s² weight = mass x g
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