Spring attached to a box; descending a hill problems

In summary, the problem involves a cyclist coasting 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. The cyclist and bicycle weigh 70 kg and assume frictional force = bv. Using the same power, the speed the cyclist can climb uphill is 3.57 m/s. The second problem involves a 0.2 kg box with a spring attached horizontally sliding on a table with coefficient of friction 0.30. A force of 21 N can compress the spring 17cm. If the spring is released, it will stretch 0.17 m beyond its equilibrium on the first swing.
  • #1
Mooshk
5
0

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?

-----------

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.
 
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  • #2
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°).
 
  • #3
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.
 
  • #4
Mooshk said:
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
Mooshk said:
mgsin3 / v = b
should be mg cos3 / v = b. Your other expressions with sin(3) are correct. Does this make sense?
 
  • #5




For the first problem, your equations and calculations seem to be correct. To find the speed the cyclist can climb uphill, you would use the same equation for power (P = Fv), but with the new force (F) that you found. This will give you the power needed to climb uphill, and then you can use the power equation again to find the corresponding speed. Your approach is correct.

For the second problem, your understanding and approach seem to be correct. You have solved for the distance the spring will stretch past equilibrium, which is the distance it will travel on the first swing. This is the total distance traveled, not just the distance past equilibrium. So your answer of 0.17 m is correct. Good job!
 

1. What is the purpose of a spring attached to a box in a hill problem?

The spring attached to a box in a hill problem is used to represent the force of gravity acting on the box as it descends the hill. It allows for the calculation of the box's acceleration and velocity.

2. How is the spring constant determined in these types of problems?

The spring constant is typically given in these types of problems or can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from equilibrium.

3. How does the angle of the hill affect the box's motion in these problems?

The angle of the hill directly affects the box's acceleration and velocity. A steeper hill will result in a greater acceleration and faster velocity, while a shallower hill will result in a slower acceleration and lower velocity.

4. What is the difference between a box sliding down a hill and a box attached to a spring sliding down a hill?

In a box sliding down a hill problem, the force of gravity is the only force acting on the box. However, in a box attached to a spring sliding down a hill problem, the force of gravity and the force of the spring are both acting on the box. This allows for a more accurate representation of real-life situations where multiple forces are involved.

5. How can these types of problems be solved mathematically?

These types of problems can be solved using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. By setting up and solving equations based on this law, the motion of the box can be determined at any point during its descent down the hill.

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