A 4.8 kg object is subjected to two forces,
F~
1 = (2.8 N) ˆı + (−2.1 N) ˆ and F~
2 =
(3.3 N) ˆı + (−10 N) ˆ. The object is at rest at
the origin at time t = 0.
What is the magnitude of the object’s acceleration?
Answer in units of m/s
2
To find the magnitude of the object's acceleration, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it can be written as:
ΣF = m * a
Where ΣF represents the sum of the forces acting on the object, m is the mass of the object, and a is the acceleration.
In this case, the object is subjected to two forces: F1 = (2.8 N) ˆı + (−2.1 N) ˆ and F2 = (3.3 N) ˆı + (−10 N) ˆ. To find the net force, we need to add these forces vectorially.
F_net = F1 + F2
= (2.8 N) ˆı + (−2.1 N) ˆ + (3.3 N) ˆı + (−10 N) ˆ
Simplifying this expression, we get:
F_net = (2.8 N + 3.3 N) ˆı + (−2.1 N - 10 N) ˆ
= 6.1 N ˆı + (-12.1 N) ˆ
Now, we can calculate the magnitude of the object's acceleration using the formula mentioned earlier:
ΣF = m * a
In this equation, ΣF is the magnitude of the net force, m is the mass of the object, and a is the magnitude of the object's acceleration.
Since the object is at rest, it means that the net force acting on it must be zero. Therefore, we can write the equation as:
0 = m * a
Now, let's substitute the known values:
0 = (4.8 kg) * a
To solve for a, we can rearrange the equation:
a = 0 / (4.8 kg)
a = 0
The magnitude of the object's acceleration is 0 m/s^2.