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A) A slideloving pig slides down a certain 40° slide in twicethe time it would t

ID: 1726970 • Letter: A

Question

A) A slideloving pig slides down a certain 40° slide in twicethe time it would take to slide down africtionless 40° slide. What is the coefficient of kineticfriction between the pig and the slide?
B) What is the smallestradius of an unbanked (flat) track around which a bicyclist cantravel if her speed is 31 km/hand the s between tires and trackis 0.31?
B) What is the smallestradius of an unbanked (flat) track around which a bicyclist cantravel if her speed is 31 km/hand the s between tires and trackis 0.31?

Explanation / Answer

The first one is rather tricky! First we need to relatethe time to go down the slide and the acceleration. From kinematics:   d = 1/2 at2          (d is constant, length of ramp)                             t2 = 2 d / a   a is proportional to 1 /t2 So if the time doubles, the acceleration decreases by a factorof 4 Without friction: a = g sin 40 = 6.30m/s2 down ramp With friction: a' = 6.30 / 4 = 1.57 m/s2 By a free body diagram, we can then use Newton's Law down theramp: ma' = mgsin40 - Ff ma' = mgsin40 -umgcos40 ugcos40 = gsin40 - a'               u = a - a' / gcos40                  = 0.63 The 2nd problem is much simpler.   If centripetalforce is supplied by friction, we can write: Ff = mv2 / r =umg                r = v2 / ug                  = (31 km/h * 1000 / 3600)2 / (0.31 * 9.8)                  = 24 m                  = (31 km/h * 1000 / 3600)2 / (0.31 * 9.8)                  = 24 m
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