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?

Question

Physics

Sammy Wise

14 February, 01:50

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?

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Christopher · Answer 1

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