4 -3 points Tipler6 7.P067. My Notes Ask Your Teacher In the figure below, the c

Question

4 -3 points Tipler6 7.P067. My Notes Ask Your Teacher In the figure below, the coefficient of kinetic friction between m13.8 kg and the shelf is 0.35. (Assume the system is initially at rest.) (a) Find the energy dissipated by friction when the block of mass m2 = 2.3 kg falls a distance y, (b) Find the change in the mechanical energy Emech of the two-block-Earth system during the time it takes the block of mass m2 falls a distance y (c) Use your result for Part (b) to find the speed of either block after the block of mass m2 falls a distance y 1.9 m. m/s

Explanation / Answer

a. While m2 falls a distance y, m1 slides that same distance. So x = y. The force of friction between m1 and the shelf is given by

Ff = mu*N = mu*m1*g = 0.35*3.8 kg*9.8 m/s^2 = 13.034 N

The work done by that friction is given by

W = Ff*x = Ff*y = 13.034y

b. The energy dissipated by friction is lost to the system. That is the only energy lost by the system.

c. The gravitational potential energy of m2 has decreased. Where that energy went is divided between the energy dissipated by friction and a gain in kinetic energy of the 2 masses.

GPE = W + KE

m2*g*y = W + KE

2.3 *9.8*y = 13.034*y + (1/2)*(m1+m2)*v^2 = 13.034*y + (1/2)*(3.8 kg + 2.3 kg)*v^2

22.54*y = 13.034*y + 3.05*v^2

3.05*v^2 = 22.54*y -13.034*y = 9.5*y = 9.5*1.9 J = 18.06

v^2 = 18.06/3.05 = 5.92

v = 2.43 m/s

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