Question

two blocks of masses 2.00kg and 3.00kg are connected by a light string that passes over a friction less pulley of moment of inertia 0.00400 kg x m2 and radius of 5.00cm. the coefficient of friction for the tabletop upon which the 3.00kg block rests is 0.300. the blocks are released from rest. using energy methods, find the speed of the upper block just as it has moved 0.600m.

Answer 1

m2 3.00 kg 2.00 kg m1 in

In the vertical axis, we have

m₁ g - T₁ = m₁a

T₁ = m₁ (g - a)                                                               { eq.1 }

In the horizontal axis, we have

T₂ - _$\mu$k m₂ g = m₂ a

T₂ = _$\mu$k m₂ g + m₂ a                                                           { eq.2 }

The net force on the pulley is given as :

F_net = T₁ - T₂$\Rightarrow$ m₁ (g - a) - (_$\mu$k m₂ g + m₂ a)

F_net = m₁ g - m₁ a - _$\mu$k m₂ g - m₂ a

F_net = (2 kg) (9.8 m/s²) - (2 kg) a - (0.3) (3 kg) (9.8 m/s²) - (3 kg) a

F_net = (10.7 N) - (5 kg) a

The torque on pulley is given as :

we know that, $\tau$ = R F   $\Leftrightarrow$ I $\alpha$ = R F

where, $\alpha$ = angular acceleration

I (a/R) = R F

(0.004 kg.m²) a = (0.05 m)² [(10.7 N) - (5 kg) a]

(0.004 kg.m²) a = (0.02675 N.m²) - (0.0125 kg.m²) a

a = (0.02675 N.m²) / (0.0165 kg.m²)

a = 1.62 m/s²

Using equation of motion 3, we have

v = v₀² + 2 a s

where, v₀ = 0

v = _$\sqrt{}$2 (1.62 m/s²) (0.6 m)

v = _$\sqrt{}$1.944 m²/s²

v = 1.4 m/s