Box and a Spring on on an inclined plane

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SUMMARY

The discussion focuses on the dynamics of a box with a mass of 30 kg on a 13-degree inclined plane, connected to a spring with a spring constant of 9 N/m. The coefficients of static and kinetic friction are both 0.12. The initial force acting on the box parallel to the ramp is calculated to be -31.76 N, indicating movement down the ramp. After sliding 0.15 meters, the net force on the box is -30.4 N, and the confusion arises regarding the spring force calculation, which is clarified as -kx where x is the displacement.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of inclined plane dynamics
  • Familiarity with spring force calculations (Hooke's Law)
  • Basic concepts of friction (static and kinetic)
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  • Study the application of Newton's second law in inclined plane problems
  • Learn about the effects of friction on motion in physics
  • Explore advanced spring dynamics and energy storage in springs
  • Investigate the role of free body diagrams in solving mechanics problems
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to forces on inclined planes and spring systems.

ac7597
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Homework Statement
A box of mass M=30 kg sits on a ramp which is tilted at angle 13 degrees. The coefficients of static and kinetic friction between the box and the ramp are both 0.12. The box is connected to a spring of spring constant k=9 Newtons per meter.

Initially, the box is held in place, and the spring is at its rest length.

The box is then released. What is the initial force on the box parallel to the ramp? Use a negative value to mean "down the ramp", and a positive value to mean "up the ramp." (Hint: draw a set of coordinate axes with X running up along the ramp, and Y running up perpendicular to the ramp).

The box starts to slide down the ramp. After it has moved a distance 0.15 meters, what is the net force on the box parallel to the ramp?

How far will the box have to slide down the ramp before the net force on it (parallel to the ramp) becomes zero? Again, use a negative value to mean "displaced down the ramp from the initial position".

At this point, when the net force on the box is zero, how will it move?
Relevant Equations
g=9.8m/s^2
Homework Statement: A box of mass M=30 kg sits on a ramp which is tilted at angle 13 degrees. The coefficients of static and kinetic friction between the box and the ramp are both 0.12. The box is connected to a spring of spring constant k=9 Newtons per meter.

Initially, the box is held in place, and the spring is at its rest length.

The box is then released. What is the initial force on the box parallel to the ramp? Use a negative value to mean "down the ramp", and a positive value to mean "up the ramp." (Hint: draw a set of coordinate axes with X running up along the ramp, and Y running up perpendicular to the ramp).

The box starts to slide down the ramp. After it has moved a distance 0.15 meters, what is the net force on the box parallel to the ramp?

How far will the box have to slide down the ramp before the net force on it (parallel to the ramp) becomes zero? Again, use a negative value to mean "displaced down the ramp from the initial position".

At this point, when the net force on the box is zero, how will it move?
Homework Equations: g=9.8m/s^2

I tried to create a free body diagram of all the forces on the box:
force x y
normal 0 Fn
gravity -mgsin(theta) -mgcos(theta)
friction uFn 0
force of spring -k(x) 0
total m(ax) m(ay)=0

thus: Fn=mgcos(theta)
m(ax)= u(mgcos(theta)) -mgsin(theta) - kx
 
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ac7597 said:
friction uFn 0
total m(ax) m(ay)=0
Please explain what you mean by the above statements.
ac7597 said:
m(ax)= u(mgcos(theta)) -mgsin(theta) - kx
So far so good. What next?
 
I set ax=0 because the first question asks for the initial force. I got kx = 31.7N but that answer is wrong.
 
ac7597 said:
I set ax=0 because the first question asks for the initial force. I got kx = 31.7N but that answer is wrong.
The initial velocity will be zero, but not the initial acceleration. The initial displacement (x) is also zero.
 
Is the initial force = u(mgcos(theta))- mgsin(theta)?
 
ac7597 said:
Is the initial force = u(mgcos(theta))- mgsin(theta)?
Yes.
 
I got -31.76N is that fine?
 
Ok -31.76N is the answer. I got the correct answer for the net force when the box move 0.15m as -30.4N. I am confused since shouldn’t the kx result in -33.11N ?
 
ac7597 said:
when the box move 0.15m
Which way? What does that make the value of x?
 
  • #10
negative... I see so its +kx
 
  • #11
ac7597 said:
negative... I see so its +kx
No, it's -kx, but since x is negative that has a positive value.
 
  • #12
thanks
 

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