Refer to the diagram shown below.
m = 8.5 kg, the mass of the cat
F = 41.0 N, the force acting up the incline on the cat
θ = 19°, the inclination of the ramp to the horizontal
u = 1.9 m/s, the initial speed along the ramp of the cat
s = 2 m, the length of the ramp
g = 9.8 m/s²
Friction is negligible.
The force F is the component of the cat's weight along the ramp.
F = mg sinθ
= (8.5 kg)*(9.8 m/s²) sin(19°)
= 27. 1198 N
The net force pushing the cat up the ramp is
41.0 - 27.1198 = 13.88 N
If the acceleration of the cat up the ramp is a, then
(8.5 kg)*(a m/s²) = 13.88 N
a = 1.6329 m/s²
Let v = the velocity at the top of the ramp.
Then
v² = u² + 2as
v² = (1.9 m/s)² + 2*(1.6329 m/s²)*(2 m) = 10.1416 (m/s)²
v = 3.185 m/s
Answer: 3.185 m/s