Forces of 10.3N north, 19.5N east, and 15.2N south are simultaneously applied to a 4.06kg mass as it rests on an air table. What is the magnitude of its acceleration?
F = 10.3i + 19.5 - 15.2i = 19.5 - 4.9i.
X = 19.5, Y = -4.9.
F = Sqrt(X^2 + Y^2)=Sqrt(19.5^2+4.9^2) =
20.1 N.
a = F/M = 20.1/4.06 = 4.95 m/s^2.
Note: The direction(angle) is not required.
To find the magnitude of the acceleration experienced by the mass, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be written as:
F = m * a
where F is the net force, m is the mass, and a is the acceleration.
In this case, we have three forces acting on the mass simultaneously: 10.3N north, 19.5N east, and 15.2N south. These forces can be broken down into their respective components along the x-axis and y-axis.
Given that the north and south directions are along the y-axis, and the east direction is along the x-axis, we can express the forces as:
Fy = 10.3N north - 15.2N south
Fx = 19.5N east
To find the net force in the x and y directions, we can sum up the individual forces in each direction:
Net Force along the y-axis (Fy):
Fy = 10.3N - 15.2N
Fy = -4.9N
Net Force along the x-axis (Fx):
Fx = 19.5N
Now, we can use these net forces to calculate the net force (F) acting on the mass:
F = sqrt(Fx^2 + Fy^2)
F = sqrt((19.5N)^2 + (-4.9N)^2)
F = sqrt(380.25N^2 + 24.01N^2)
F = sqrt(404.26N^2)
F ≈ 20.11N
Now, we can substitute the values of the net force (F) and the mass (m) into the equation F = m * a to find the acceleration (a):
20.11N = 4.06kg * a
a = 20.11N / 4.06kg
a ≈ 4.95 m/s^2
Therefore, the magnitude of the acceleration of the mass is approximately 4.95 m/s^2.
Fnet = F1 + F2+ . . .
Fnet =(10.3 N [North])+(19.5 N [South]) - 15.2 N [South] = 14.6 N [North]
Fnet = ma
a = Fnet/m
14.6 N [North]/(4.06 kg)
a = 3.60 m/s/s [North]