Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

Question

Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is: (A) 3.5 (B) 3.3 (C) 3.0 (D) 2.5

Tardigrade Admin · Accepted Answer

Considering the equilibrium of A, we get 10 a =10 g sin 30°-T - mN where N=10 g cos 30° ∴ 10 a=(10/2) g-T-μ × 10 g cos 30° but a=0, T =mB g 0=5 g-mB g-(0.2 √3/2) × 10 g ⇒ mB=3.268 ≈ 3.3 kg

Q. Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

3.5

3.3

3.0

2.5

Considering the equilibrium of A, we get 10a=10gsin30∘−T - mN where N=10gcos30∘ ∴10a=210​g−T−μ×10gcos30∘ but a=0,T=mB​g 0=5g−mB​g−20.23​​×10g ⇒mB​=3.268≈3.3kg

Q. Two blocks are connected over a massless pulley as shown in fig. The mass of block $A$ is $10 k g$ and the coefficient of kinetic friction is $0.2$ . Block A slides down the incline at constant speed. The mass of block $B$ in $k g$ is: