# Ap Physics C 1989 M2?

##### 1 Answer

a) This is a basic application of newton's second law.

#2Mg - T_v = 2Ma#

#2Mg - 2Ma = T_v#

#2M(g - a) = T_v#

b) First of all, we know that what will make the pulley have a non zero angular acceleration is the tension.

We know torque to be

#T_vR - T_hR = I(a/r)#

#T_vR - T_hR = 3MR^2(a/R)#

#T_v - T_h = 3Ma#

#2gM - 2Ma - T_h = 3Ma#

#2gM - 5Ma = T_h#

#2gM-5Ma = T_h#

c) Now for some linear dynamics! We know that the force of friction is given by

Our expression for net force will therefore be

#T_h - F_f = 3Ma#

#2Mg - 5Ma - mu m_"total" g = 3Ma#

Here we will be taking the mass of the top block for the normal force , because we are considering the friction between the two blocks.

#2Mg -5Ma - mu 4Mg = 3Ma#

#2Mg - 4Mgmu = 8Ma#

#2Mg - 4Mgmu = 8Ma#

#g - 2gmu = 4a#

We know the value of

#g- 2gmu = 4a#

#g(1 - 2mu) = 8#

#1 - 2mu = 8/g#

#mu ~~ 0.1#

d) The only force acting on the top box is friction, therefore

#4Mg mu = M_c a_c#

#4Mg mu = 4Ma_c#

#g(0.1) = a_c#

#a_c ~~ 1# m/#s^2#

This concludes this problem. Hopefully this helps!