Answer:
2.32 m/s
Step-by-step explanation:
If an 8.90 kg block of ice slides down a 1.15 m frictionless ramp to reach a speed of 2.87 m/s, you want to know its final speed if there were friction opposing the motion with a force of 11.0 N.
Energy
The kinetic energy at the bottom of the frictionless ramp is ...
KE = 1/2mv²
KE = 1/2(8.90 kg)(2.87 m/s)² = 36.654205 J
Friction
When friction is introduced, the work done to oppose the friction is ...
W = Fd
W = (11 N)(1.15 m) = 12.65 J
Hence the remaining energy of the block at the bottom of the ramp with friction is ...
KE' = 36.654205 -12.65 J = 24.004205 J
This corresponds to a speed of ...
v = √(2·KE/m) = √(2·24.004205/8.9) ≈ 2.32 . . . . m/s
The speed at the bottom of the ramp with friction is about 2.32 m/s.
__
Additional comment
We can find the slope of the ramp by equating the ending kinetic energy to the beginning potential energy. As you can see, that is not relevant to the problem, since the opposing force is parallel to the ramp.