Two forces act on a 16 kg object. the first force has a magnitude of 68 n and is directed 24 degrees north of east. the second force is 32n , 48 degrees of west. what is the acceleration of the object resulting from the application of these forces? a. 1.6 m/s^2, 5.5 degrees north of west b. 1.9 m/s^2, 18 degrees north of west c. 2.4 m/s^2 , 34 degrees north of east d. 3.6 m/s^2 , 5.5 degrees north of west e.4.1 m/s^2, 52 degrees north of east

e. 4.1 m/s^2, 52 degrees north of east

Explanation:

Two forces act on a 16 kg object. The first force has a magnitude of 68 N and is directed 24 degrees north of east. The second force is 32N , 48 degrees of west. What is the acceleration of the object resulting from the application of these forces? a. 1.6 m/s^2, 5.5 degrees north of west b. 1.9 m/s^2, 18 degrees north of west c. 2.4 m/s^2 , 34 degrees north of east d. 3.6 m/s^2 , 5.5 degrees north of west e.4.1 m/s^2, 52 degrees north of east

There are two components of force

finding the sum of the horizontal component of the forces

Efx

 68 cos 24 - 32cos48 = 40.7088 N

Resolve the sum of the vertical components  of the forces

    68 sin 24   +  32 Sin 48 = 51.438  

now we find the resultant of the two components

R=(Fy^2+Fx^2)^0.5

      resultant force =   (51.438 ^ 2  +  40.7088 ^ 2)^ 1/2  

    = 65.597

recall that force is the product of mass and acceleration

ma = 65.597

m=16 kg

 a = 65.597/16 = 4.09 m/s/s

tan=EFy/EFx

    tan  = 51.438/ 40.7088 =

1.263

taking the tan inverse of both sides

 = 51.64 degree north of east

the area between the graph and the t axis