Question

Two blocks of masses m1=1kg and m2=2kg are connected by a non deformed light spring.they are lying on a rough horizontal surface.the coefficient of friction between the blocks and the surface is 0.4.What minimum constant force F has to be applied in horizontal direction to the block of mass m1 in order to shift the other block?

Answer 1

Given that,

Mass of block $M_{1}=1\ kg$

$M_{2}=2\ kg$

Coefficient of friction = 0.4

We need to calculate the minimum force

Using balance equation for M₂

$F_{min} x-\dfrac{1}{2}k'x^2-km_{2}gx=0$

$F_{min} -\dfrac{1}{2}k'x-km_{2}g=0$....(I)

Using balance equation for M₁

$k'x=km_{1}g$....(II)

Now, put the value of k'x in equation (I)

$F_{min}=\dfrac{1}{2}km_{1}g+km_{2}g$

$F_{min}=kg(\dfrac{m_{1}}{2}+m_{2})$

Put the value into the formula

$F_{min}=0.4\times9.8(\dfrac{1}{2}+2)$

$F_{min}=9.8\ N$

Hence, The minimum force is 9.8 N.