Question

A block of mass m1 = 0.500 kg sits on a frictionless surface and is connected by a weightless string to a weight of mass m2 = 0.200 kg that hangs from a pulley. The system is initially at rest. If the mass m2 is released and drops for 1.00 m, what is the speed of the system? Assume that mass m1 does not reach the edge of the surface. Use energy considerations, not force considerations. What is the speed of the system, if the surface is rough and has a kinetic friction coefficient k = 0.250?

Answer 1

Let, the initial and final heights of the mass m₂ be h_i and h_f respectively.

Hence, ( h_i - h_f ) = 1 m.

Hence, distance moved by the mass m₁ is also d = 1 m.

Since, m₁g is the normal reaction force by the surface on m₁,

frictional force on the mass m₁ = F_f = _k x m₁g.

Using the principle of conservation of energy, we get :

Initial P.E. of m₂ = Final P.E. of m₂ + Work done by F_f + Total K.E. of the system of mass ( m₁ + m₂ )

or, m₂gh_i = m₂gh_f + F_f x d + ( m₁ + m₂ ) v² / 2

or, m₂g x ( h_i - h_f ) = _k x m₁g x d + ( m₁ + m₂ ) v² / 2

or, ( m₁ + m₂ ) v² / 2 = m₂g x ( h_i - h_f ) - _k x m₁g x d

or, ( 0.5 + 0.2 ) x v² / 2 = 0.2 x 9.8 x 1 - 0.25 x 0.5 x 9.8 x 1

or, 0.35 x v² = 0.735

or, v² = 0.735 / 0.35 ~ 2.1

or, v = 2.1 ~ 1.45.

Hence,  the speed of the system is : 1.45 m / s.