a skier of mass m is standing on a slope with an angle of elevation ø. After be
ID: 2105637 • Letter: A
Question
a skier of mass m is standing on a slope with an angle of elevation ø. After being given a tiny nudge, the skier slides down the slope, grabs a rope at the bottom of the slope, and swings up to a height h. The coefficient of kinetic friction between the skier and the slope is µk.
a) what is the minimum coefficient of static friction required so that the skier is initially at rest at the top of the slope without using his poles?
b) how much energy is lost while the skier slides down the incline?
c) what is the coefficient of kinetic friction?
Explanation / Answer
FOLLOW THIS METHID
fsmax= µs*Fn
µs=fsmax/Fn
where Fn is the normal force
101/(140*9.81)
ANS µs =0.0735 (No units because its a coefficient)
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3) You need to draw pictures to help when you start doing problems like this as its a geometry problem.
You need to resolve the force due to gravity for two separate directions.
Direction one is the force acting parallel to the slope.
(68*9.81) sin 30 = 333.54 N pushing the skier down the slope.
The second force is the "normal force" the force that acts perpendicular to the slope.
(68*9.81) sin(60) = 577.71N
fsmax= µ*Fn
fsmax= 0.18*577.71
fsmax= 103.99N
Resultant force = 333.54N(the force acting parallel to the slope)-103.99N(the resistive force) =
229.55 N
F=ma
a=F/m
= 229.55/68
ANS= 3.376m/s^2 (3dp) Answer value seems reasonable.
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6)....OUCH. LOL. X component is dependant on flight time. Flight time is dependant on the vertical velocity.
Vertical Velocity = 27 sin(30)
= 13.5m/s in the Y direction
Uhm.... t = 2*V/g
= (2*13.5)/9.81
= 2.75 Seconds (flight time)
Horizontal distance = s*t
= 27cos(30)* 2.75
= 64.30m
Vertical Distance = t/2 * 13.5 (vertical velocity)
(symetrical arc due to no resistance, so half total flight time must be height travelled)
2.75/2 * 13.5=18.56m
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7)
a=V^2/r
a=29^2/57
Ans = a=14.75m/s^2
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b)
Centripetal force = (m*V^2)/r
=ERROR I need mass of car to continue. ( I think)
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8)
S= 1/2a t^2
t^2= S/2a
t^2=1.25/2*9.81
ANS t=0.25 seconds
Speed= distance/time
= 0.415/0.25
ANS = 1.66 m/s
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9)
F=(Gm)/r
G=6.672*10^-11 Nm^2/kg^2
F= (6.672*10^-11 * 71020)/ 500*10^3
Ans = 9.477*10^-12N I have know idea if this is correct as the numbers are meaningless to me but gravity is a weak force and the asteroid only has a mass of 71 tons so its plausible
Weight = mg
= 74 * 9.477*10^-12
= 701.29*10^-12 N
Ans = 701.29 pN
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10)
This is the same as question 7 only this time the car has a mass horaar.
Speed= Distance/time
distance= 2 * Pi*r
= 2* Pi*50
= 314.16 m
Speed = 314.16/14.3(s)
Speed = 21.969 m/s
a=V^2/r
21.969^2/50
Ans a= 9.653 m/s^2
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Centripetal force = (m*V^2)/r
= 600* 21.969^2/50
Ans = 5791.64N
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