Well, I must say, your cat seems to be on quite an uphill journey. Let's calculate her speed at the top of the incline, shall we?
First, we need to break down the forces acting on Ms. Along the inclined plane, there are two forces: the force you applied (50.0N) and the gravitational force pulling her down the ramp.
The force of gravity can be calculated using the formula: F_gravity = mass * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, F_gravity = 9.00 kg * 9.8 m/s^2 = 88.2 N.
Now, let's resolve the force parallel to the ramp (50.0N) into its x and y components. The component of the force parallel to the ramp is given by F_parallel = force * sin(theta), where theta is the angle of inclination (19.0 degrees in this case). Therefore, F_parallel = 50.0N * sin(19.0) ≈ 16.4N.
Since there are no other forces acting on the cat in the horizontal direction, the net force (F_net) acting on Ms. is given by F_net = F_parallel = 16.4N (since the gravitational force is perpendicular to the motion).
Now, we can use Newton's second law in the horizontal direction, which states that F_net = mass * acceleration. Rearranging the formula, we have acceleration = F_net / mass = 16.4N / 9.00 kg ≈ 1.822 m/s^2.
Finally, we can use the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (2.00 m/s at the bottom), a is the acceleration (1.822 m/s^2), and s is the distance traveled (2.00 m on the ramp).
Plugging in the values, we get v^2 = (2.00 m/s)^2 + 2 * 1.822 m/s^2 * 2.00 m ≈ 11.288. Taking the square root, we find v ≈ 3.359 m/s.
So, when your cat Ms. reaches the top of the incline, she'll be zooming along at approximately 3.359 m/s. Time to give her a round of applause for conquering that frictionless ramp!