Question

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?

Answer 1

`2.289 ms^-2`

Explanation:

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`