Question 1161028
NOTE: If you have to "show your work" you may want to use the number of decimal places/significant figures that would make your teacher/instructor/professor happy.
I tend to use more decimal places than needed during intermediate calculations and round at the end, because that is what is required in my line of work.
 
a) Assuming a horizontal, level floor, the normal force (perpendicular to the surface, pressing the box into the floor) is
{{{(70kg)*(9.807ms^9-2)=686.49N}}}
The horizontal friction force opposing the movement is
{{{0.50*(686.49N)=343.245N}}}
The net force on the box is
{{{400N-343.245N=56.755N}}}
The box will accelerate in the direction of the 400N force.
The acceleration will be
{{{56.755N/"70 kg"=highlight(0.81ms^(-2))}}}
 
b) {{{drawing(500,350,-.5,9.5,-1,6,
red(line(0,0,8.66,5)),rectangle(0,0,8.66,5),
arrow(0,0,-1,0),arrow(8.66,0,11,0),
line(0,0,8.66,0),line(0,0,0,5),
arrow(0,0,0,5),arrow(3.66,0,8.66,0),
red(arrow(4.33,2.5,8.66,5)),
rectangle(0,0,-.2,0.2),locate(4.33,2.5,red(F=400N)),
locate(4,0.15,F[x]),locate(0.1,2.8,F[y]),
green(arc(0,0,4,4,330,360)),locate(1.3,0.7,green(30^o))
)}}}
 
{{{F[x]=(400N)*cos(30^o)=(400N)*0.8660=highlight(346.4N)}}}
{{{F[y]=(400N)*sin(30^o)=(400N)*0.5000=highlight(200N)}}}

 
c) Now the normal force is
{{{686.49N-F[y]=686.49N-200N=486.49N}}}
That will make the friction force opposing the movement
{{{(486.49N)*0.50=243.245N}}}
The net horizontal force acting on the box will be
{{{346.4N-243.245N=103.155N}}}
Acting on a mass of 70kg, that net horizontal force will produce a horizontal acceleration of
{{{103.155N/"70 kg"=highlight(1.47ms^(-2))}}}