Box and a Spring on on an inclined plane

  • #1
ac7597
126
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
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,798
8,196
friction uFn 0
total m(ax) m(ay)=0
Please explain what you mean by the above statements.
m(ax)= u(mgcos(theta)) -mgsin(theta) - kx
So far so good. What next?
 
  • #3
ac7597
126
6
I set ax=0 because the first question asks for the initial force. I got kx = 31.7N but that answer is wrong.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,798
8,196
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.
 
  • #5
ac7597
126
6
Is the initial force = u(mgcos(theta))- mgsin(theta)?
 
  • #7
ac7597
126
6
I got -31.76N is that fine?
 
  • #8
ac7597
126
6
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 ?
 
  • #10
ac7597
126
6
negative... I see so its +kx
 
  • #11
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,798
8,196
negative... I see so its +kx
No, it's -kx, but since x is negative that has a positive value.
 
  • #12
ac7597
126
6
thanks
 

Suggested for: Box and a Spring on on an inclined plane

Replies
10
Views
518
Replies
24
Views
412
  • Last Post
Replies
27
Views
3K
Replies
76
Views
3K
Replies
11
Views
86
  • Last Post
Replies
6
Views
300
Replies
15
Views
429
Replies
19
Views
368
Top