Two horizontal forces act on a 1.9 kg chopping block that can slide over a frictionless kitchen counter, which

Question

lies in an xy plane. One force is F1 = (3.3 N) + (4.5 N) . Find the acceleration of the chopping block in unit-vector notation for each of the following second forces.
(a) F2 = (-3.3 N) + (-4.5 N)

Answer 1

Lets calculate the angle between them .

cos(theta)=F1.F2/IF1IXIF2I
=(3.3*-3.3)+(4.5*-4.5)/(squreroot(3.3^2)+(4.5)^2Xsqureroot(-3.3)^+(4.5)^2
cos(theta)=-1
Theta=180 degrees.

Resultant force acting on it = 0
Therefore No Accelertation.
acceleration=(0)+(0)

Answer 2

As presented, the force F1 is equal and opposite to the force F2 so the net force is zero and the acceleration is zero. The presentation is sloppy, however, and one must assume that you meant to say there were x and y components of each of the two forces ( not summing them unless you add unit vectors in the appropriate directions).

Answer 3

Acceleration of the block is a=3.0m/s^2i . Mass of the block is m=2.0 kg Resultant stress that quickens the block is Fr = ma = 2 * 3i = 6N i Fr = F1 + F2 F2 = Fr - F1 = (6 i + 0 j) - (3 i + 4 j) = (6 - 3) i + (0 - 4) j = 3 i - 4 j

Answer 4

Here the forces are equal and opposite to each other.
Resultant force = 0
acceleration a = F/m =0