Box and a Spring on on an inclined plane

Click For Summary

Homework Help Overview

The problem involves a box on an inclined plane connected to a spring, with specific parameters such as mass, angle of inclination, and coefficients of friction. Participants are exploring the forces acting on the box as it begins to slide down the ramp and how these forces change with displacement.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the box, including gravitational, frictional, and spring forces. There is an exploration of free body diagrams and the equations of motion. Questions arise regarding the initial force calculations and the implications of setting acceleration to zero.

Discussion Status

There is active engagement with various interpretations of the forces involved. Some participants have provided calculations for the initial force and net force after a displacement, while others express confusion about the signs and values of these forces. Guidance has been offered regarding the correct interpretation of the spring force in relation to displacement.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on understanding the relationships between the forces and the motion of the box.

ac7597
Messages
126
Reaction score
6
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
 
Physics news on Phys.org
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
 

Similar threads

Replies
6
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
4K
Replies
2
Views
674
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
3K
Replies
16
Views
2K
Replies
4
Views
2K