ANSWER
191.5 N
EXPLANATION
First, let's draw a free-body diagram of the forces acting on the safe,
By Newton's second law, we have that,
[tex]\begin{gathered} F_N+F_{ay}-F_g=0 \\ and \\ F_{ax}-F_f=m\cdot a_{} \end{gathered}[/tex]
We know that the safe is moving to the right at a constant speed, which means that the acceleration is zero,
[tex]F_{ax}-F_f=0[/tex]
Solving for the force of friction,
[tex]F_f=F_{ax}[/tex]
The x-component of the applied force is,
[tex]F_{ax}=F_a\cdot\cos 40\text{\degree}=250N\cdot\cos 40\text{\degree}\approx191.5N[/tex]
Hence, the force of friction between the floor and the safe is 191.5 N.
