Spring attached to a box; descending a hill problems

[Mooshk]

Homework Statement

First Problem:
A cyclist coasts down a 3 degree hill at a constant speed of 2.2 m/s. When the cyclist pumps hard, they can descend the hill at 12.6 m/s. Using the same power, what is the speed the cyclist can climb uphill? The cyclist and bicycle weigh 70 kg. Assume that frictional force = bv.

Second Problem:
A 0.2 kg box with a spring attached horizontally slides on a table where the coefficient of friction is 0.30. A force of 21 N can compress the spring 17cm. If the spring is released from this position, how far beyond its equilibrium will it stretch on the first swing?

Homework Equations

Force(friction) = bv
Power = W/t; Power = Force(v)
F(spring) = kx; k = F/x
Gravitational Potential Energy = mgh
Kinetic Energy = 1/2 mv^2
Elastic Potential Energy = 1/2 kx^2
Work(Friction) = change in total energy
(Based on my guesses at solving these problems)

The Attempt at a Solution

For the first problem:

mgsin3 / v = b
(70)(9.8)sin3 / 2.2 = b
16.32 = b

mgsin3 - bv + F(cyclist) = 0
35.90 - 205.63 + F = 0
169.73 N = F

P = F(cyclist) * v
P = 2138.60 W

Then, for going uphill,
-mgsin3 - bv + F = 0 ? Then I would solve for F again, and sub that into P=FV to find speed?

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For the second problem:
I don't think I understand this correctly. If anyone can explain this problem to me, I'd appreciate it.

k = F/x
k = 21/0.17
k = 123.52

I think the initial and final kinetic energies are 0...

The initial potential elastic energy
E = 1/2 kx^2
E = 1/2 (123.53)(0.17 ^2)
E = 1.785 J

Force (friction) = (uk)mg
Force = (0.3)(0.2)(9.8)
Force = 0.588 N

Work (friction) = F(friction) * distance (or the change in total energy... I think I looked at this wrong)
Work = 0.588 N * distance

So the final elastic energy is
E = 1/2 k(distance ^2)
E = 1/2 (123.53) (distance ^ 2)

So then if I equate the two elastic energies:

1.785 J = 1/2 (123.53) (distance ^ 2)
3.57 = 123.53 (distance ^2)
0.0289 = (distance ^2)
0.17 m = distance

Now I'm not sure if I solved for the total distance traveled, or just past equilibrium.

I don't know if I understood these problems correctly. I'll appreciate any help you can give me. Thanks in advance.

 

[Undoubtedly0]

Welcome to Physics Forums! For the first problem, I noticed you used sin(3°) as part of your normal force calculation. Look over your trigonometry again; I think it is supposed to be cos(3°).

 

[Mooshk]

Thanks for the reply.
Can you explain why the force from the cyclist would be cos?
I thought that the force of gravity assisting the cyclist is parallel to the incline, that's why I used sin.

I don't know how to edit the first post, but I found the correct solution to the second problem.

 

[Undoubtedly0]

Thanks for the reply.
Can you explain why the force from the cyclist would be cos?
I thought that the force of gravity assisting the cyclist is parallel to the incline, that's why I used sin.

You're welcome! You are correct in that the force of gravity assisting the cyclist is parallel to the incline. However, cosine must be used when calculating the normal (and thus frictional) force - in other words, your expression

mgsin3 / v = b

should be mg cos3 / v = b. Your other expressions with sin(3) are correct. Does this make sense?