Answer
(a) The pushing force required is $446.88N$.
(b) The pushing force required is $241.08N$.
Work Step by Step
As the crate does not move in $y$ direction, the normal force acting on the floor equals the weight of the crate.
$$F_N=m_{crate}g=60\times9.8=588N$$
(a) To overcome static friction, the pushing force $F_p$ has to surpass the maximum static frictional force $f_s^{max}$. In other words, the required force is $$F_p=f_s^{max}=\mu_sF_N=0.76\times588=446.88N$$
(b) To slide the crate at a constant speed with no acceleration means that the pushing force equals the kinetic frictional force:$$\sum F=0$$ $$F_p-f_k=0$$ $$F_p=f_k=\mu_k\times F_N=0.41\times588=241.08N$$
