. ... A mass m is pushed against a spring of spring constant k = 12000 N/m. The
ID: 2113085 • Letter: #
Question
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A mass m is pushed against a spring of spring constant k = 12000 N/m. The mass slides on a horizontal smooth surface except for a rough one meter long section, and then it encounters a loop of radius R as shown in the figure. The mass just barely makes it around the loop without losing contact with the track (i.e. the normal force at the top of the loop is zero). The coefficient of kinetic friction (mu k) between the mass and the rough surface is 0.35, m = 2 kg and R = 1 m. calculate speed of the mass at the top of loop calculate energy of the mass last due to friction calculate distance spring the spring was compassed.Explanation / Answer
Let the Velocity at Top point be v
As it is Barely Complete the loop
Therefore at the Topmost Point
Centerifugal Force = Weight of the Mass
mv^2/r = mg
Therefore
v = sqrt(rg)
= sqrt(1*9.8)
= 3.13 m/sec
Energy Lost Due to Friction = Work Done against Friction
= ukmgs
= 0.35*2*9.8*1
= 6.86 J
Let by x Distance the Spring Compressed
Therefore
Kinetic Energy at the Top Point + Potential Energy at Top point = Elastic potential Energy of the Spring - Work Done against Friction
Therefore
0.5*2*9.8 + 2*9.8*2 = 0.5*12000*x^2 - 0.35*2*9.8*1
x = 0.09649 m = 9.649 cm
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