In the figure below, the coefficient of kinetic friction between m1 = 3.6 kg and

ID: 1623508 • Letter: I

Question

In the figure below, the coefficient of kinetic friction between m1 = 3.6 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 = 1.9 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

a) Energy dissipated = Frictional force x distance covered = 0.35(3.6) (9.8) (y ) = 12.348 y J ( since y is not provided , I have left y as it is)

b) Change in mechanical energy = energy lost to friction = - 12.348 y J

c)change in energy = final E - initial E

- 12.348 Y = 1/2 m1v^2 + 1/2 m2v^2 + m2g (Y) ( where Y is the distance through both the masses move)

1.9 ( - 12.348 - 18.62) = 1/2 v^2 ( 5.5)

v = 4.62 m/s apprx ( ignoring the -ve sign)