1.) A 0.580- k g wood block is firmly attached to a very light horizontal spring
ID: 2253959 • Letter: 1
Question
1.)A 0.580-
kg wood block is firmly attached to a very light horizontal spring (k=170N/m) . It is noted that the block-spring system, when compressed 5.3cm and released, stretches out 2.6cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?
2.)A bicyclist coasts down a 7.0
? hill at a steady speed of 8.0m/s Assuming a total mass of 95
kg (bicycle plus rider), what must be the cyclist's power output to climb the same hill at the same speed? I keep getting the answer to be 907.68W but apparently that's wrong.
3.) A 1200
kg car has a maximum power output of 130hp . How steep a hill can it climb at a constant speed of 55
km/h if the frictional forces add up to 670N ?
I would really appreciate correct answers with work to get them.
Explanation / Answer
(a) we know, at max Compression & Elongation
velocity=0
change in potential energy=energy consumed by friction
1/2k(0.053)^2 - 1/2k(0.026)^2 = coeff. *mg*(0.053 +0.026)
coeff.= 0.405
(b)
(c)1hp=746 W ,let hill be of angle $
we know P=F.v
F=mgsin$ + friction [ for climbing up]
F=1200*9.8*sin$ + 670=11760sin$ + 670
v=55 km/h=15.28m/s
P=F.v
130*746 W= (11760sin$ +670)*15.68 W
sin$=0.483
$=28.88 degree
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