F=ma, so a = F/m = 10N/3Kg = 3.3m/s/s

NOTE: If you prefer to see the units cancel, write Kg*m/s/s instead of N.
 
 

What force will accelerate a 6000Kg truck at .5m/s/s?

F= ma = 6000Kg*.5m/s/s =3000N
 
 

Two skaters each apply 30N of force to a .15Kg hockey puck at once.  The angle between the forces is 30°. Find the accelration provided by the skaters AND the angle of the puck's motion.
STEP 1: Make a sketch... I decided to put one vector on the X axis so that I don't have to worry about any Y component for it.

STEP 2: Get the net force by adding the vector's X and Y components...
V1x = 30N

V2x = 30N * cos30 = 26N
V2y = 30N * sin30 = 15N

do you see why placing one vector on an axis makes your life easier? anyway...

NET X = 30 + 26 = 56N
NET Y = 15N

NET FORCE= sqrt(56² + 15²) = 58N

STEP 3: Use 2nd law...

a = F/m = 58N/.15Kg = 386.7m/s/s

STEP 4: Find angle of net force...

q = inv tan(15N/56N) = 15°, directly between the two forces