As per the given question,
mass of the block A(m1)=12kg
Coefficient of kinetic friction (μ)=0.14
Angle of inclination (θ)=32.0∘
Block A is slides with constant speed in downward direction,
now, let the tension in the string be T,and mass of the block B is m2
So,T=m2g−−−−−−−(i)
m1gsinθ−T−μm1gcosθ=0
T=m1gsinθ−μm1gcosθ−−−−−−(ii)
From equation (i) and (ii)
m2g=m1gsinθ−μm1gcosθ
⇒m2=gm1gsinθ−μm1gcosθ
⇒m2=m1sinθ−μm1cosθ
⇒m2=12sin32−12×0.14cos32
⇒m2=4.93kg
So, mass of block B will be 4.93 kg
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