Figure shows tow blocks in contact sliding down an inclined surface of inclination 30^0. The friction coefficient between the block o fmass 2.0 kg and the incline is mu_1 and that between the block of mas 4.0 kg and the incline is mu_2. Calculate teh acceleration of the 2.0 kg bloc if a. mu_1=0.20 and mu_2 =0.30 b. mu_1=0.30 and mu_2=0.20. Take g=10m/s^2.

Figure shows tow blocks in contact sliding down an inclined surface of

1.	Figure shows tow blocks in contact sliding down an inclined surface of inclination 30^0. The friction coefficient between the block o fmass 2.0 kg and the incline is mu_1 and that between the block of mas 4.0 kg and the incline is mu_2. Calculate teh acceleration of the 2.0 kg bloc if a. mu_1=0.20 and mu_2 =0.30 b. mu_1=0.30 and mu_2=0.20. Take g=10m/s^2.
Answer» Solution :From the free body diagram R=4g cos 30^0` `rarrR=4xx10xx sqrt3/2` `20sqrt3` `mu_2R+4a-p-4g sin30^0` `rarr 0.3(40)cos30^0+4a-p-40sin20^0` =20 `R_1=2g cos 30^0 = 10 ........iii sqrt3` `p+2a-mu_1R_1-2g SIN 30^0=0`....iv ` From equation ii. `6sqrt3+4a-p-20=0` From equation iv `p+2a+2sqrt3-10=10` `6sqrt+6a+30+2sqrt3=0` `rarr6a=30-8sqrt3` `=30-13.85=1615` `rarr a=16.15/6` `=2.69=2.7m/s^2` B. In this CASE 4 kg block will travel with more acceleration because, coefficient of friction is less than that of 2 kg. So, they will MOVE separately. Drawin the free body diagram of 2 jkg mass only, it can be found that `a=2.4m/s^2`