Spring and tension problem

In summary, the conversation discusses an ideal spring system with two blocks and a pulley. The forces on each block are analyzed in equilibrium and the tension in the string is calculated. The force constant of the spring is also determined. The string is then cut and the time for the 4.0kg block to hit the floor is calculated, along with the frequency of oscillation and maximum speed attained by the 8.0kg block. The formula for frequency is also reviewed and used to determine the frequency in this system.
  • #1
aeroengphys
21
0

Homework Statement


An ideal spring of unstretched length .20m is placed horizontally on a frictionless table as shown above. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0kg. The 8.0kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m = 4.0kg hangs from the other end of the spring. When this spring-and-blocks system is in equilibrium, the length of the spring is .25m and the 4.0kg block is .70m above the floor.

(a) On the figures below, draw free-body diagrams showing and labeling the forces on each block when the system is in equilibrium.

(b) Calculate the tension in the string.

(c) Calculate the Force constant of the spring

**The string is now cut at point P, a given location on the horizontal frictionless surface between block M and the pulley. **

(d) calculate the time taken by the 4.0 kg block to hit the floor.

(e) Calculate the frequency of oscillation of the 8.0kg block.

(f) Calculate the meximum speed attained by the 8.0kg block.


Homework Equations


net F = 0
f = kx
y = 1/2gt^2
f = cycles/sec


The Attempt at a Solution



a) For the 4kg block, i said there was an mg downward and a tension upward.
For the 8kg block, I said there was an mg downward, a normal force upwards, a tension to the right, and a spring force to the left.

b) F = 0
W = T
40N = T

c) F spring = T
kx = 40N
k(.25) = 40N
k = 160 N/s

d) y = 1/2gt^2
.7m = 1/2(9.8m/s^2)(t^2)
t = .374s

e) i know that the formula is f = cycles/sec but I'm a little confused on how exactly i can determine the time for a cycle.

f) PE spring = KE block
1/2kx^2 = 1/2mv^2
1/2(160N/s)(.25m)^2 = 1/2(8kg)v^2
v = 1.12 m/s


If you could give me some help with e, as well as check over my answers briefly, i'd greatly appreciate it.
 
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  • #2
aeroengphys said:
a) For the 4kg block, i said there was an mg downward and a tension upward.
For the 8kg block, I said there was an mg downward, a normal force upwards, a tension to the right, and a spring force to the left.
Good.

b) F = 0
W = T
40N = T
Good. (You are using g = 10 m/s^2 here.)

c) F spring = T
kx = 40N
k(.25) = 40N
k = 160 N/s
Careful! In Hooke's law, F = kx, x is the amount the spring is stretched, not the length of the spring.

d) y = 1/2gt^2
.7m = 1/2(9.8m/s^2)(t^2)
t = .374s
Good. Here you are using g = 9.8 m/s^2. (Recheck that third digit.)

e) i know that the formula is f = cycles/sec but I'm a little confused on how exactly i can determine the time for a cycle.
That's just the definition of frequency, not how to find it. What you need to understand is simple harmonic motion with springs, and how to calculate the frequency of a mass oscillating on a spring. See: http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1"

f) PE spring = KE block
1/2kx^2 = 1/2mv^2
1/2(160N/s)(.25m)^2 = 1/2(8kg)v^2
v = 1.12 m/s
You'll need to redo this with the correct value for spring stretch.
 
Last edited by a moderator:
  • #3
Thanks for you help Doc. I fixed the value for the spring constant and got 800N/s. I also reviewed my notes on springs, and found that period is equal to 2(pi)√m/k That would mean T = 2(pi)√8kg/800N/m = .6283s -> f = 1.59 Hz. Does that sound right?
 
Last edited:
  • #4
That sounds about right. My only comment would be to make up your mind as to which value you are using for g; I would use 9.8 m/s^2, not 10.
 

What is a "Spring and Tension Problem"?

A "Spring and Tension Problem" is a physics problem that involves calculating the force of a spring as it is stretched or compressed, and the tension force in a string or rod that is attached to the spring.

What are the key concepts involved in solving a "Spring and Tension Problem"?

The key concepts involved in solving a "Spring and Tension Problem" include Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed; equilibrium, which means that the forces acting on the system must balance out; and the concept of tension, which is the force that is transmitted through a string or rod when it is pulled tight.

How do you calculate the force of a spring in a "Spring and Tension Problem"?

The force of a spring can be calculated using Hooke's Law, which states that the force (F) is equal to the spring constant (k) multiplied by the distance the spring is stretched or compressed (x). This can be represented by the equation F = -kx, where the negative sign indicates that the force is in the opposite direction of the displacement of the spring.

What is the tension force in a "Spring and Tension Problem"?

The tension force in a "Spring and Tension Problem" is the force that is transmitted through a string or rod when it is pulled tight. In these types of problems, the tension force is often used to balance out the force of the spring, creating an equilibrium state.

What are some real-life applications of "Spring and Tension Problems"?

"Spring and Tension Problems" have many real-life applications, including in engineering, such as designing bridges and buildings, and in everyday items like car suspension systems and door hinges. These problems are also commonly used in sports, such as calculating the force of a tennis racket string or a bowstring in archery.

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