The force of Friction

Question

!!f_1:!! friction between 2Kg and 4Kg block

!!f_2:!! friction between 4Kg block and ground

Solution

Free body diagram,

For block 2Kg

!!f_1= mu_1m_1g= 0.25 xx 2 xx10= 5N!!

!!10- f_1= m_1a_1!!

!!=> 10-5 = (2)a_1!!

!!=> a_1 = 2.5 m//s^2!!

For block 4Kg

!!f_2(max)= mu_2 (m_1+m_2)g= 0.1xx6xx10=6N!!

If !!f_2= 6N!!, then

!!f_1-f_2 = m_2a_2!!

!!5-6 = (4)a_2!!

!!=> a_2= - 1/4!!.

Direction of net force will be in oppsoite direction of !!a_2!!, which is not possible.

So !!f_2(max)<= f_1!! therefore !!f_2= 5N!!

!!f_1-f_2= m_2a_2!!

!!=> a_2= 0 m//s^2!!

Option (c) is correct.

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