Al I. In the above figure, the coefficient of kinetic friction between mi and th

Question

Al I. In the above figure, the coefficient of kinetic friction between mi and the inclined plane is = 0.50, = 30° and 'ni = 10 kg. In Homework #4 you found the value of m2 that allows mi to slide down the inclined plane with constant velocity. a) What is the work done by friction when mi slides 1 meter down the plane? Now assume the mass of mi is doubled while m2 remains the same. If mi is released from rest, find b) the work done against friction when mi slides 1 meter down the plane c) the change in kinetic energy of the system after mi has slid 1 meter down the plane d) the change in potential energy of the system after mi has slid 1 meter down the plane e) the relationship between the quantities calculated in b, c, d that follows from the Work-Energy Theorem and/or conservation of energy. You may assume the cord and the pulley have negligible masses.

Explanation / Answer

a] Work by friction = -uk m1g cos theta d

= -0.50*10*9.8* cos 30 degree *1 = -42.4 J answer

b] work by friction = -uk m1g cos theta d

= -0.50*20*9.8* cos 30 degree *1 = -84.8 J answer

c] Lets find what the value of m2 was. m1g sin 30 degree - uk m1g cos theta = m2g

or 10*0.5 - 0.5*10*0.866 = m2. or m2 = 0.67 kg

Change in KE of system = work done by all forces = -84.8 + m1g sin theta *d - m2g d

= -84.8 + 20*9.8*0.5 - 0.67*9.8*1

= 6.634 J

d] Change in PE = -(m1g sin theta *d - m2g d)

= -( 20*9.8*0.5 - 0.67*9.8*1 ) = -91.434 J answer

e] Work by friction = change in KE + change in PE

b = c+d , can easily be seen

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