Question

Your cat Goldie (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 39.930 m long and inclined 27.0 ∘ above the horizontal. Since Goldie can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 37.40 N force parallel to the ramp (there is a set of stairs alongside the frictionless ramp on which you can walk). If Goldie is moving at 2.000 m/s at the bottom of the ramp (assume she has gotten a running start), what is her speed when she reaches the top of the incline?

Answer 1

The speed when se reaches the top of the incline is 0.28 m/s

Step-by-step explanation:

The work done is equal to the change of kinetic energy, then:

Wg + Wf + Wn = ΔEk

Where

Wg = work done by gravity

Wf = work done by force

Wn = work done by normal force

`-mgdsin\theta +Fd+0=(1)/(2) *m*(v_(2) ^(2) -v_(f) )`

Where

m = 8.5 kg

g = 9.8 m/s²

d = 39.93 m

F = 37.4 N

vf = 2 m/s

Replacing:

`39.93*(-8.5*9.8*sin27+34.4)=(1)/(2) *8.5*(v_(2) ^(2) -2^(2) )\\v_(2) =0.28m/s`