**Problem : **
A 10 kg mass, initially at rest, experiences three forces: one North
with magnitude 10 N, one East, with magnitude 20 N and one Northeast
with magnitude 30 N. Find the resulting acceleration. After 10 seconds,
assuming the forces continue to act while the object is in motion, what
is the object's velocity? How far has it traveled?

We solve the problem by drawing a free body diagram:

Solution 1

Now we find the sum of the

*x* and

*y* components of all three forces:

*F*_{x} = 20 + 30 cos 45 = 41.2

*F*_{y} = 10 + 30 sin 45 = 31.2

And sum these two vectors. The magnitude of the resultant force is
given by:

And the direction is given by:

*θ* = tan^{-1}(31.2/41.2) = 37.1^{o}, North of East.
Thus the object experiences a force of 51.7 Newtons, directed

37.1^{o} North of East.
Now we must find its acceleration, using Newton's Second Law:

*a* =

=

= 5.17

, 37.1

^{o} North of East

To find the final velocity and position of the object, we simply use
the equations learned in

kinematics:

*v*_{f} = *v*_{o} + *at* = 0 + (5.17)(10) = 51.7 m/s

*x*_{f} = *x*_{o} + *v*_{o}*t* + (1/2)*at*^{2} = 0 + 0 + (1/2)(5.17)(10)^{2} = 258.5 m

Thus, after 10 seconds, the object is moving with a velocity of 51.7
m/s, directed
31.7 degrees North of East. Also, the object has moved 258.5 meters, in
the same
direction.