Sam (80 k g ) takes off up a 50- m -high, 10 %u2218 frictionless slope on his je
ID: 2117014 • Letter: S
Question
Sam (80kg ) takes off up a 50-m-high, 10%u2218 frictionless slope on his jet-powered skis. The skis have a thrust of 220N . He keeps his skis tilted at 10%u2218after becoming airborne, as shown in the figure.How far does Sam land from the base of the cliff?
A 3.10g coin is placed 19.0cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of %u03BCs = 0.700 and %u03BCk = 0.590. What is the maximum angular velocity with which the turntable can spin without the coin sliding?
Explanation / Answer
A) velocity at cliff is V
Work done by upthrust force and gravity = F*S - mgh = 220*50/Sin10 - 80*9.8*50 = 24.146 KJ
0.5*m*V^2 = 24146
=> V = Sqrt(2*24146/80) = 24.57 m/s
Projectile motion:
x = VCos10 *t
y = VSin10 * t - 0.5g*t^2
y at landing = -50
=> -50 = 4.266t - 4.9t^2
=> t = 3.66s
x = VCos10*3.66 = 88.56 m from the cliff
B)Normal reaction on coin = N
N = mg
friction = (coefficient of static friction)*N = 0.7*mg
centripital force must not exceed static friction
=> m*w^2*R = 0.7*mg
=> w = Sqrt(0.7*9.8/0.19) = 6.01 rad /s
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