In a pulley system, $m_{1}$ $m_1$ is $2.0 k g$ $2.0kg$ and $m_{2}$ $m_2$ is $1.0 k g$ $1.0kg$ . The coefficient of kinetic friction between $m_{1}$ $m_1$ and the table is $μ_{k} = 0.15$ $mu_k=0.15$ . What is the acceleration of the pulley system?

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In a pulley system, $m_{1}$ $m_1$ is $2.0 k g$ $2.0kg$ and $m_{2}$ $m_2$ is $1.0 k g$ $1.0kg$ . The coefficient of kinetic friction between $m_{1}$ $m_1$ and the table is $μ_{k} = 0.15$ $mu_k=0.15$ . What is the acceleration of the pulley system?

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Answer 1 · 2016-06-17T14:53:50

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

Explanation:

given given

GIVEN

$m_{1} = 2 k g, m_{2} = 1 k g$ $m_1=2 kg , m_2 =1kg$
$Coefficient of kinetic friction between m_{1} and table (μ_{k}) = 0.15$ $"Coefficient of kinetic friction between " m_1 " and table" (mu_k)=0.15$
$Let$ $"Let"$
$T be tension on string$ $color (blue) T " be tension on string"$
$and a be the acceleration of the system$ $" and " color(green) a" be the acceleration of the system "$

Now considering the forces on $m_{1}$ $m_1$
Normal reaction $N = m_{1} g$ $N = m_1g$
Resisting force of kinetic friction $f_{k} = μ_{k} \times N = μ_{k} m_{1} g$ $f_k=mu_kxxN=mu_km_1g$
Gravitational pull on $m_{1}$ $m_1$ being vertical to T it will not creat any resistance.

So for $m_{1}$ $m_1$ 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

$m_2g-mu_km_1g=m_1a+m_2a$

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

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

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

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In a pulley system, $m_{1}$ $m_1$ is $2.0 k g$ $2.0kg$ and $m_{2}$ $m_2$ is $1.0 k g$ $1.0kg$ . The coefficient of kinetic friction between $m_{1}$ $m_1$ and the table is $μ_{k} = 0.15$ $mu_k=0.15$ . What is the acceleration of the pulley system?

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Explanation:

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In a pulley system, m_1m1m_1 is 2.0kg2.0kg2.0kg and m_2m2m_2 is 1.0kg1.0kg1.0kg. The coefficient of kinetic friction between m_1m1m_1 and the table is mu_k=0.15μk=0.15mu_k=0.15. What is the acceleration of the pulley system?

1 Answer

Explanation:

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In a pulley system, $m_{1}$ $m_1$ is $2.0 k g$ $2.0kg$ and $m_{2}$ $m_2$ is $1.0 k g$ $1.0kg$ . The coefficient of kinetic friction between $m_{1}$ $m_1$ and the table is $μ_{k} = 0.15$ $mu_k=0.15$ . What is the acceleration of the pulley system?