• Support PF! Buy your school textbooks, materials and every day products Here!

Dynamics/ Simple Harmonic motion help

  • Thread starter SpIIIdD
  • Start date
  • #1
2
0
Hello everyone, i'm new here.
I'll just post my question- hopefully someone can answer.

A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 n/m. The pulley is frictionless. The block is released from rest when the spring is un-stretched. The block moves 20 cm down th incline before coming to rest. Find the coefficient of friction between the block and incline.


m= 2 kg
d (delta x)= 20 cm (or 0.2 m)
k= 100 N/m
angle to the horizontal, theta= 37 deg.
coef of Fk= ?




a relevant equation, i think, for this problem is coef of Fk= Fk/Fn (normal force) but i'm not 100% sure about anything else.



So first i tried to draw the full body diagram of the forces of that are acting on the block.
The mg force points straight down, the kinetic frictional force should point diagonally to the left and the normal force points diagonally upward (parallel to the surface of contact where the block rests).
However, i'm not exactly sure what the force going diagonally to the right and downward- opposite to the force of kinetic friciton going opposite the direction at movement. I thought maybe that the force would be the force exerted by the string, Fx, but when i conceptually try to calculate it (with the formula Fx= -kx or kx ) i seem to obtain 0 (since the spring is being uncompressed and therefore returning to equilibrium, k should equal 0? (and multiply then by x- 0.2- and the answer is zero or have i got it wrong)?

I'm just generally confused as to how to continue with this ques and am wondering maybe there's something worng with my reasoning. I know that what i've done should be folowed up by solving a system of equations to get the final value but i'm not sure how to go about doing it.

Help?
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,028
2,416
The mg force points straight down, the kinetic frictional force should point diagonally to the left
I hope you mean parallel to the incline.
and the normal force points diagonally upward (parallel to the surface of contact where the block rests).
You mean perpendicular to the surface of contact. That's what "normal" means.
However, i'm not exactly sure what the force going diagonally to the right and downward- opposite to the force of kinetic friciton going opposite the direction at movement.
What force would that be?

Anyway, I think it would be easier to tackle this problem by using the work-energy theorem or total (not just mechanical) energy conservation.
 
  • #3
2
0
I hope you mean parallel to the incline.

You mean perpendicular to the surface of contact. That's what "normal" means.

What force would that be?

Anyway, I think it would be easier to tackle this problem by using the work-energy theorem or total (not just mechanical) energy conservation.
ok, so i drew two different free- body diagrams- one for the block statically in position and another one for it in movement. I derived the formula to determine thr coef of kinetic friction: mewk= Fk/FN. I'm not exactly sure how to apply the work energy theorem to solve this problem since i'm not looking for work and there isn't the coef of kinetic friction in the formula (?). Particularly i'm not sure how to utilize the spring constant and the known displacement to solve the problem.
Can someone give me a further hint?
 
  • #4
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,028
2,416
The work-energy theorem can be used for things other than just finding work. Don't forget that friction does work therefore μk appears in the work done by friction, so if you apply the work-energy theorem you will have an equation with μk in it.

What does the work-energy theorem say? What is an expression for it? How many forces do work and what are the corresponding expressions?
 
Last edited:

Related Threads on Dynamics/ Simple Harmonic motion help

  • Last Post
Replies
1
Views
2K
Replies
3
Views
741
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
942
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
1
Views
1K
Replies
8
Views
7K
Replies
1
Views
2K
Replies
12
Views
1K
Top