Source: (Various old midterms. and Fall 2010 assignment question) A 0.600 Kg blo

Question

Source: (Various old midterms. and Fall 2010 assignment question) A 0.600 Kg block is held against a compressed spring of stiffness 28.0 N/m and compression 1.05 m on a 22.62' incline that is 2.60 m long. The coefficient of kinetic friction between the block and inclined is 0.400 The external force holding the block against the compressed spring is then removed, and the compressed spring pushes the block up the incline from rest. (The block is not attached to the spring so it is unable to push on the block after it reaches its equilibrium length.) What is the work done by gravity on the block along the length of the incline? What is the work done by kinetic friction along the length of the incline? With what speed docs the block leave the incline?

Explanation / Answer

mass of the block m = 0.6 kg spring constant k = 28 N / m compression x = 1.05 m angle of inclination = 22.62 degrees length of the incline L = 2.6 m coefficient of kinetic friction between block and the inclined surface = 0.4 (a). work done by gravity on the block along the length of the incline W = -mg sin * L              W = - 5.88 J (b). work done by kinetic friction along the length of the incline W ' = -[ mg cos ]* L             W ' = -5.6448 J (c). applied force on the block F = kx = 29.4 N work done by F is W " = FL                                      = 76.44 J net work done w = W + W '+W"                             = 64.9152 J from work energy theorem ,w = change in KE                                               = ( 1/ 2) m[v^ 2-u^ 2] where u = initial velocity of the block = 0 So, w = ( 1/ 2) mv^ 2 from this required speed v = [2w / m]                                           = 14.7 m / s

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