Question

This Physics uses Algebra and Trigonometry

Answer 1

(1) Consider the tension in the string is T
Now from FBD of mass . M = 12 kg block
Mg - T = Ma
T = Mg - Ma ----------(1)
Now considering the FBD of mass , m = 10 kg
T - mgSin27 = ma
T = ma + mgSin27 ----------(2)
Equating 1 and 2
Mg - Ma = ma + mgSin27
a(M+m) = Mg -mgSin27
a(12+10) = 12*9.81 - (10*9.81Sin27)
a = 3.326 m/s²
(2) Now in second case
Considering the FBD of M
Mg - T = Ma
T = Mg - Ma -----------(1)
Now considerin the FBD of mass m
T - mgSin27 - f_k = ma
Where f_k is friction force = u_k*R
R = mgCos27 , therefore
f_k = u_k*mgCos27
Now putting the value in above equation
T - mgsin27 - (u_k*mgCos27) = ma
T = ma +mgSin27 + (u_k*mgCos27) -----------(2)
Equating 1 and 2
Mg - Ma = ma +mgSin27 + (u_k*mgCos27)
a(M+m) = Mg - [mgSin27 + (u_k*mgCos27)]
a(12+10) = 12*9.81 - [(10*9.81*Sin27)+(0.24*10*9.81*cos27)]
a(12+10)= 117.72 - [44.536 + 20.98 ]
a(12+10) = 52.204
a = 2.373 m/s²