Answer:
The correct Answer to the question is : e) 4.1 m/s^2, 52 degrees north of east
Explanation:
F1= 68 N < 24º = 62.12 i + 27.65 j
F2= 32N < 132º = -21.41 i + 23.78 j
R= F1+F2= 40.71 i +51.43 j = 65.59 N < 51.63 º
By 2nd law of newton:
F= m * a
R= m*a
a= R/m
a= 4.1 m/s² < 52º (52 degrees north of east)
I consideer 0º at the EAST axis.
Answer:
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[tex]\alpha[/tex]=EFy/EFx
tan = 51.438/ 40.7088 =
1.263
taking the tan inverse of both sides
[tex]\alpha[/tex] = 51.64 degree north of east
