Question 7 = 16.6 Submit My Answers Give Up A 8.00-kg block of ice, released fro

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Question 7 = 16.6 Submit My Answers Give Up A 8.00-kg block of ice, released from rest at the top of a 1.35-m-long frictionless ramp, slides downhill, reaching a speed of 2.75 m/s at the bottom. Correct Significant Figures Feedback: Your answer 16.61 ° was either rounded differently or used a different number of significant figures than required for this part. Part B What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.2 N parallel to the surface of the ramp? 4.12 m/s Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining; no points deducted Provide Feedback Continue

Explanation / Answer

using kinematic equations
vf² = vi² + 2ad
a = (vf² - vi²) / 2d

a = (2.75² - 0²) / 2(1.35)
a = 2.80 m/s²

for a sloped frictionless surface
a = gsin
sin = a/g

sin = 2.80 / 9.81
= 16.58°

a 10.2 N force acting on an 8 kg mass creates an acceleration of
a = F/m
a = 10.2/8
a = 1.275 m/s²

subtracting this from our frictionless acceleration leaves
2.80 - 1.275 = 1.525 m/s²

vf² = vi² + 2ad
vf² = 0² + 2(1.525)(1.35)
vf² = 4.1175
vf = 2.02 m/s

OR if you are working energy equations

KE = PE
½mv² = mgh
½v² = gh
v²/2g = h
v²/2g = dsin
sin = v²/2gd
sin = 2.75²/2(9.81)1.35
= 16.58°

KE = PE - U
½mv² = mgh - Fd
v² = 2(mgh - Fd) / m
v² = 2(8.0(9.81)(1.35sin16.58) - 10(1.35)) / 8.0
v² = 2.57
v = 1.60 m/s

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