Help with a physics problem

Question

Answer 1

Let

a be the common acceleration of the two blocks.

(Their accelerations have different direction, but common magnitude a.)

b.

Block 1:

T₁ - m₁gμ = m₁a, where

m₁gμ is the force of friction,

T₁ - m₁gμ is the net force acting on block 1.

Block 2:

m₂g - T₂ = m₂a, where

m₂g is the weight of the block,

m₂g - T₂ is the net force acting on block 2.

Pully:

(T₂ - T₁)r = Mr²a/r, or simplified

T₂ - T₁ = Ma, where

T₂ - T₁ is the net force acting on the pully,

(T₂ - T₁)r is the torque acting on the pully,

Mr² is the moment of inertia of the pully,

a/r is angular acceleration of the pully.

c., d.

We have three equations and three unknowns T₁, T₂, a

T₁ - m₁gμ = m₁a

m₂g - T₂ = m₂a

T₂ - T₁ = Ma

a = (m₂g - m₁gμ)/(m₁ + m₂ + M) = g(m₂ - m₁μ)/(m₁ + m₂ + M) = gQ

where we abbreviate Q = (m₂ - m₁μ)/(m₁ + m₂ + M)

T₁ = m₁gQ + m₁gμ = m₁g(Q + μ)

T₂ = m₂g - m₂gQ = m₂g(1 - Q)

e.

The time t it takes to drop height h with acceleration a is

t = √(2h/a)

In that time the speed becomes

v = at = a√(2h/a) = √(2hgQ)

f.

Starting from rest, at the beginning the kinetic energy of the whole system is 0.

Let the potential energy of block 2 at the beginning be also 0.

Then after dropping height h

-m₂gh is the potential energy of block 2

m₂v²/2 is the kinetic energy of block 2

m₁v²/2 is the kinetic energy of block 1

Mr²(v/r)²/2 is the kinetic energy of the pully, which can be simplified into

Mv²/2

The total energy plus the work performed by the force of friction over the distance h must equal initial energy, i.e., 0:

0 = Mv²/2 + m₁v²/2 + m₂v²/2 - m₂gh + m₁gμh

g.

v = √(2h(m₂g - m₁gμ)/(m₁ + m₂ + M) = √(2hgQ)

h.

Q = (8 - 4×0.4)/(4 + 8 + 2) = 0.457

a = 9.81×0.457 = 4.48 m/s²

T₁ = 4×9.81×(0.457 + 0.4) = 33.6 N

T₂ = 8×9.81×(1 - 0.457) = 42.6

v = √(2×1.5×4.48) = 3.67 m/s