Learning goal: to practice problem-solving strategy 6.1 work and kinetic energy. your cat "ms." (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined 19.0 ∘ above the horizontal. since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 41.0 n force parallel to the ramp. if ms. is moving at 1.90 m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline?

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

3.185 m/s

She sold 135 peaches, because if you multiply 2.50 eith 135, you'd get 337.5, the 0 can be added or not.