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

B, D

Explanation

From the free body diagram
R=4g cos 30^0`
`rarrR=4xx10xx sqrt3/2`
`20sqrt3`

`mu_2R+4a-p-4g sin 30^0`
`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`