A 4.6 kg object is subjected to two forces,
F~1 = (2.8 N)iˆ+(−3.5 N)jˆ and
F~2 = (4 N)iˆ+(−11.5 N)jˆ. The object is at rest at the origin at time t = 0.What is the magnitude of the objects acceleration?
-i've tried using the equation netforce=M*A but have had no success.
x direction
Fx = 2.8 + 4 = 6.8
Fy = -3.5 - 11.5 = - 15
Ax = 6.8/4.6
Ay = -15/4.6
|A| = sqrt (Ax^2 + Ay^2)
To find the magnitude of the object's acceleration, you need to determine the net force acting on the object and then use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Here's how you can calculate the net force:
1. Add the x-components and y-components of the individual forces to get the net force in each direction.
Net Force in the x-direction (F~x):
F~x = F~1x + F~2x = (2.8 N) + (4 N) = 6.8 N
Net Force in the y-direction (F~y):
F~y = F~1y + F~2y = (-3.5 N) + (-11.5 N) = -15 N
2. Calculate the magnitude of the net force.
Magnitude of the net force (|F~|):
|F~| = √(F~x^2 + F~y^2)
|F~| = √((6.8 N)^2 + (-15 N)^2)
|F~| = √(46.24 N^2 + 225 N^2)
|F~| = √(271.24 N^2)
|F~| ≈ 16.5 N
Now that you have calculated the net force acting on the object, you can use Newton's second law to find the acceleration:
Acceleration (a) = Net force (|F~|) / Mass (M)
a = |F~| / M
a = 16.5 N / 4.6 kg
a ≈ 3.59 m/s^2
Therefore, the magnitude of the object's acceleration is approximately 3.59 m/s^2.