On the horizontal surface of a truck a block of mass 2 kg is placed ($\mu=0.5$ static and kinetic) a

Question

On the horizontal surface of a truck a block of mass 2 kg is placed ($\mu=0.5$ static and kinetic) and the truck is moving with an acceleration of $6 \mathrm{~m} / \mathrm{s}^{2}$. Then the frictional force on the block will be (Take $g=10 \mathrm{~m/s}^2$)

Professor PC Thomas Classes · Answer

Limiting static friction is $f_{L} = \mu_s N = 0.5 	imes 2 	imes 10 = 10 \mathrm{~N}$. The pseudo force required to accelerate the block with the truck is $F = m a = 2 	imes 6 = 12 \mathrm{~N}$. Since $F > f_L$, the block will slip. Thus, the frictional force acting is kinetic friction (assumed equal to static here), which is $f_k = \mu_k N = 0.5 	imes 20 = 10 \mathrm{~N}$.