Find the acceleration reached by each of the two objects shown if the coefficient of kinetic friction between the 7.00-kg object and the plane is 0.250.
To find the acceleration of the two objects, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma).
Let's consider the 7.00-kg object first. It experiences two forces: the force of gravity (mg) and the force of kinetic friction (fk). The force of gravity is given by the equation Fg = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s²).
Fg = mg = (7.00 kg)(9.8 m/s²) = 68.6 N
The force of kinetic friction can be determined using the equation fk = μkN, where μk is the coefficient of kinetic friction. N is the normal force acting on the object, which is the component of the force of gravity perpendicular to the plane.
N = mgcosθ, where θ is the angle between the plane and the horizontal.
Since the plane is not inclined in the given problem, θ = 0°, and cos(0°) = 1. Therefore, N = mg.
fk = μkN = μkmg = (0.250)(7.00 kg)(9.8 m/s²) = 17.15 N
Since the force of kinetic friction acts in the opposite direction to the motion, it is negative. So the net force acting on the object is:
Fnet = Fg - fk = 68.6 N - 17.15 N = 51.45 N
Using Newton's second law, we can find the acceleration:
Fnet = ma
51.45 N = (7.00 kg) * a
a = 51.45 N / 7.00 kg ≈ 7.36 m/s²
Therefore, the acceleration reached by the 7.00-kg object is approximately 7.36 m/s².
To find the acceleration of the second object, we need more information. Please provide details or specify what the second object is.
This is an example of why some questions go unanswered.
There are no diagrams or graphics or images on this web site.
Unfortunately, this is not clearly explained when students post their questions and attempt to include figures.