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A 3kg block is placed over a 10kg block ...

A 3kg block is placed over a 10kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.2. IF a horizontal force of 20N is applied to 3kg block. Find the acceleration of the two blocks.

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Here , mass of upper block `m_1=3kg`
Mass of lower block `m_2=10kg `
horizontal force `F=20N`
Coefficient of friction between the blocks
`mu=0.2`
Let `a_1 and a_2` be accelerations of upper and lower blocks respectively then
the equation of motion of upper block is
`F-f=m_1a_1`
`a_1= (F-f)/m_1= (F-mu m_1 g)/m_1`
`=(20 N -(0.2)(3kg) (10 ms^-2))/(3 kg)`
`=(20 N-6N)/(6 kg)=14/3 ms^-2`

The equation of motion of lower block is `f=m_2a_2`

`a_2=f/m_2= (mum_1 g)/m_2`
`=((0.2)(3kg) (10ms^-2))/(10 kg)=(6N)/(10 kg)=0.6 ms^-2`
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