In a pulley system, #m_1# is #2.0kg# and #m_2# is #1.0kg#. The coefficient of kinetic friction between #m_1# and the table is #mu_k=0.15#. What is the acceleration of the pulley system?

1 Answer
Jun 17, 2016

Answer:

#2.289 ms^-2#

Explanation:

given

GIVEN

  • #m_1=2 kg , m_2 =1kg#
  • #"Coefficient of kinetic friction between " m_1 " and table" (mu_k)=0.15#

  • # "Let"#
    #color (blue) T " be tension on string"#
    #" and " color(green) a" be the acceleration of the system "#

Now considering the forces on #m_1#
Normal reaction # N = m_1g#
Resisting force of kinetic friction #f_k=mu_kxxN=mu_km_1g#
Gravitational pull on #m_1# being vertical to T it will not creat any resistance.

So for #m_1# we have

#T-mu_km_1g=m_1a......(1)#

Considering the forces on #m_2# we can write

#m_2g-T=m_2a.....(2)#

Adding equation (1) and equation(2) we get

#m_2g-mu_km_1g=m_1a+m_2a#

#=>a (m_1+m_2)=(m_2-mu_km_1)g#

#=>a =((m_2-mu_km_1)g)/(m_1+m_2)=((1-0.15xx2)xx9.81)/(1+2)#

#=(0.7xx9.81)/3=2.289ms^-2#