The given problem can be exemplified in the following diagram:
In the diagram we have the following forces:
[tex]\begin{gathered} F=\text{ pushing force} \\ F_x=\text{ x-component of the force} \\ F_y=\text{ y-component of the force} \\ N=\text{ normal force} \\ F_f=\text{ friction force} \\ W=\text{ weight} \end{gathered}[/tex]Now we will determine the value of the acceleration of the box. To do that we will add the forces in the horizontal direction:
[tex]\Sigma F_h=F_x-F_f[/tex]According to Newton's second law we have that the sum of forces must be equal to the product of the mass and the acceleration:
[tex]F_x-F_f=ma[/tex]To determine the value of the friction force we use the following relationship:
[tex]F_f=\mu N[/tex]To determine the value of the normal force we add the forces in the vertical direction:
[tex]\Sigma F_v=N-F_y-W[/tex]Since there is no movement in the vertical direction we have:
[tex]N-F_y-W=0[/tex]Now we solve for the normal force:
[tex]N=F_y+W[/tex]To determine the vertical component of the force we use the right triangle shown in the diagram. We use the function sine and we get:
[tex]\sin 35.7=\frac{F_y}{F}[/tex]Now we multiply both sides by "F":
[tex]F\sin 35.7=F_y[/tex]Now we substitute this value in the formula for the normal force:
[tex]N=F\sin 35.7+W[/tex]Now we substitute this value in the formula for the friction force:
[tex]F_f=\mu(F\sin 35.7+W)[/tex]Now we substitute this value in the formula for the horizontal forces:
[tex]F_x-\mu(F\sin 35.7+W)=ma[/tex]Now, to determine the horizontal component of the force we use the function cosine:
[tex]\cos 35.7=\frac{F_x}{F}[/tex]Now we multiply both sides by "F":
[tex]F\cos 35.7=F_x[/tex]Now we substitute in the sum of horizontal forces:
[tex]F\cos 35.7-\mu(F\sin 35.7+W)=ma[/tex]Now we solve for "a" by dividing both sides by "m":
[tex]\frac{F\cos 35.7-\mu(F\sin 35.7+W)}{m}=a[/tex]To determine the mass we use the formula for the weight:
[tex]W=mg[/tex]Where:
[tex]g=\text{ acceleration of gravity}[/tex]Now we divide both sides by "g";
[tex]\frac{W}{g}=m[/tex]Now we substitute the values:
[tex]\frac{(401N)\cos 35.7-(0.587)(401N\sin 35.7+245N)}{\frac{245N}{9.8\frac{m}{s^2}}}=a[/tex]Solving the operations we get:
[tex]1.779\frac{m}{s^2}=a[/tex]Now, we will determine the final velocity of the movement. We will use the following equation of motion:
[tex]2ad=v^2_f-v^2_0[/tex]Since the box starts from rest this means that the initial velocity is zero:
[tex]2ad=v^2_f[/tex]Now we take the square root to both sides:
[tex]\sqrt[]{2ad}=v_f[/tex]Now we substitute the values:
[tex]\sqrt[]{2(1.779\frac{m}{s^2})(3.37m)}=v_f[/tex]Solving the operations:
[tex]3.46\frac{m}{s}=v_f[/tex]Now we use the following equation of motion to determine the time:
[tex]v_f=v_0+at[/tex]Since the initial velocity is zero:
[tex]v_f=at[/tex]Now we divide by the acceleration:
[tex]\frac{v_f}{a}=t[/tex]Substituting the values:
[tex]\frac{3.46\frac{m}{s}}{1.779\frac{m}{s^2}}=t[/tex]Solving the operations:
[tex]1.94s=t[/tex]Therefore, it takes the box 1.94 seconds to move.
