# 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?

Jun 17, 2016

$2.289 m {s}^{-} 2$

#### Explanation:

GIVEN

• ${m}_{1} = 2 k g , {m}_{2} = 1 k g$
• $\text{Coefficient of kinetic friction between " m_1 " and table} \left({\mu}_{k}\right) = 0.15$

• $\text{Let}$
$\textcolor{b l u e}{T} \text{ be tension on string}$
$\text{ and " color(green) a" be the acceleration of the system }$

Now considering the forces on ${m}_{1}$
Normal reaction $N = {m}_{1} g$
Resisting force of kinetic friction ${f}_{k} = {\mu}_{k} \times N = {\mu}_{k} {m}_{1} g$
Gravitational pull on ${m}_{1}$ being vertical to T it will not creat any resistance.

So for ${m}_{1}$ we have

$T - {\mu}_{k} {m}_{1} g = {m}_{1} a \ldots \ldots \left(1\right)$

Considering the forces on ${m}_{2}$ we can write

${m}_{2} g - T = {m}_{2} a \ldots . . \left(2\right)$

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

${m}_{2} g - {\mu}_{k} {m}_{1} g = {m}_{1} a + {m}_{2} a$

$\implies a \left({m}_{1} + {m}_{2}\right) = \left({m}_{2} - {\mu}_{k} {m}_{1}\right) g$

$\implies a = \frac{\left({m}_{2} - {\mu}_{k} {m}_{1}\right) g}{{m}_{1} + {m}_{2}} = \frac{\left(1 - 0.15 \times 2\right) \times 9.81}{1 + 2}$

$= \frac{0.7 \times 9.81}{3} = 2.289 m {s}^{-} 2$