Can anyone please help me with this spring question?

  • #1
I am not looking for an answer, just a method that gives me a result close to 1871.88. I have tried this for hours and when I entered that value it says it is wrong but by a small amount. Please, can someone help me as soon as possible, the assignment is due at 9pm tonight.

1. Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.

The question: Calculate the maximum force constant of the spring kmax that can be used in order to meet the design criteria

Homework Equations


Work=Force*distance
Kinetic Energy=0.5*mass*velocity^2
Work=change in KE
Force=K(spring constant)*x(distance)
0.5*mass*velocity^2 = 0.5*force*x^2 (distance)
F=kx=mgsin(theta)+friction

The Attempt at a Solution


F=kx=mgsin(theta)+friction
F=1490*sin(21)+533=1066.97

0.5*m*v^2=0.5*Fx
*2
m*v^2=Fx

m=1490/9.8=152.04

x=mv^2/F
x=152.04*2^2/1066.97

x=0.57

k=F/x

k= 1066.97/0.57= 1871.88
 

Answers and Replies

  • #2
PeroK
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I am not looking for an answer, just a method that gives me a result close to 1871.88. I have tried this for hours and when I entered that value it says it is wrong but by a small amount. Please, can someone help me as soon as possible, the assignment is due at 9pm tonight.

1. Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.

The question: Calculate the maximum force constant of the spring kmax that can be used in order to meet the design criteria

Homework Equations


Work=Force*distance
Kinetic Energy=0.5*mass*velocity^2
Work=change in KE
Force=K(spring constant)*x(distance)
0.5*mass*velocity^2 = 0.5*force*x^2 (distance)
F=kx=mgsin(theta)+friction

The Attempt at a Solution


F=kx=mgsin(theta)+friction
F=1490*sin(21)+533=1066.97

0.5*m*v^2=0.5*Fx
*2
m*v^2=Fx

m=1490/9.8=152.04

x=mv^2/F
x=152.04*2^2/1066.97

x=0.57

k=F/x

k= 1066.97/0.57= 1871.88

It's really difficult to follow what you are trying to do when you don't describe things. The first calculation looks good but you haven't said what you are doing. Here's how I would have written that:

When the crate comes to rest, the force of the spring must not be more than the effective weight of the crate plus static friction (otherwise the crate will rebound). Therefore:

Then your calculation. Although, it should actually be a ##\le## not ##=##.

After that I can't figure out what you're doing. Is it something to do with conservation of energy?

If you describe what you're doing it makes it so much easier to help.
 
  • #3
epenguin
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F=kx=mgsin(theta)+friction

Shouldn't that + be a -? Opposing forces.
 
  • #4
Ok, thank you for helping me first of all. Basically what Im trying to do is figure out the spring constant by obtaining the force, and the distance the spring moved and the force exerted on the spring
First I have tried finding the force on the spring by calculating force on the slope, as the crate is moving down the slope, so I have calculate the slope force+ friction, as those are the forces allowing the crate to move.

Then as kinetic energy equals work done, and work done also equals the spring energy, I have made them equal each other as they both equal work done. Spring energy can equal either 1/2* F (spring force)*x (distance) or 1/2 k (spring constant)*x^2 (spring distance moved). Where I have used the first equation.

I then rearranged the kinetic energy and spring energy for x, which gave me

x=152.04*2^2/1066.97

x=0.57

Which I then used to calculate spring constant by:

k=Force/x (distance spring moved)

k= 1066.97/0.57= 1871.88
 
  • #5
PeroK
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Ok, thank you for helping me first of all. Basically what Im trying to do is figure out the spring constant by obtaining the force, and the distance the spring moved and the force exerted on the spring
First I have tried finding the force on the spring by calculating force on the slope, as the crate is moving down the slope, so I have calculate the slope force+ friction, as those are the forces allowing the crate to move.

Then as kinetic energy equals work done, and work done also equals the spring energy, I have made them equal each other as they both equal work done. Spring energy can equal either 1/2* F (spring force)*x (distance) or 1/2 k (spring constant)*x^2 (spring distance moved). Where I have used the first equation.

I then rearranged the kinetic energy and spring energy for x, which gave me

x=152.04*2^2/1066.97

x=0.57

Which I then used to calculate spring constant by:

k=Force/x (distance spring moved)

k= 1066.97/0.57= 1871.88

Actually, I just worked it out. You've taken KE and spring energy into account, but you've forgotten two other forms of energy!
 
  • #6
Actually, I just worked it out. You've taken KE and spring energy into account, but you've forgotten two other forms of energy!

Can you please tell me what I am missing, I think one of them is gravitational potential energy, but I'm not sure about the other
 
  • #7
PeroK
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F=kx=mgsin(theta)+friction

Should t that + be a -? Opposing forces.

That equation is all right (although it should actually be an inequality).

You also need ##F + F_f > mgsin(\theta)## in order for the block to stop. But that must be the case as soon as the block starts to slow down.
 
  • #8
PeroK
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Can you please tell me what I am missing, I think one of them is gravitational potential energy, but I'm not sure about the other

Friction does work!
 
  • #9
Ok, I think I know what you are trying to tell me, and I think I may have done the method you are trying to tell me, but I always end up stuck in the last part, with the rearranging:

force on spring= force on slope+friction
F=kx=mgsin(theta)+friction
F=1066.97

Elastic spring energy= kinetic energy+ gratitational potential energy - work done by friction
0.5*k*x^2=0.5*m*v^2+mgh-friction*distance
0.5kx^2=304.08+11128.28 - 4264
0.5kx^2=7168.36

*2

k(1066.97/k)^2= 14336.72

I got k in the brackets by doing F=kx, which rearranging gives x=1066.97/k

I really appreciate the help, guys, and I'm sure I'm nearly there, but in the meantime can you please still keep helping me until I get the answer, thank you.
 
  • #10
Actually, I just worked it out. You've taken KE and spring energy into account, but you've forgotten two other forms of energy!

can you please tell me, what you got, and how. I know you are probably busy, but I would really appreciate it if you could help me some more.
 
  • #11
Or what value it is closer to ou of;
1871
1873

or is it neither of those two, and it is my decimal figures which are wrong
 
  • #12
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Assuming you did it correctly, you know kx2 and you know kx, so you have two equations in two unknowns.
 
  • #13
From my calculations I have got:

kx=1066.97
kx^2=14336.72

Which if I then both equal to k, I get

1066.97/x=14336.72/x^2

Which then I multiply with x, to eliminate one side, and then rearrange to get;

x=14336.72/1066.97
x=13.44

Which is the wrong answer as x cannot be higher than 8, if 8 includes 8, and if I then put in F=kx, and rearrange for k, I get 79.39, which is absolutely the wrong number. I have no idea where I'm messing up, and I really need help.
 
  • #14
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Your change in potential energy is incorrect. What distance does the crate drop vertically?

Chet
 
  • #15
I'm not really sure, how do I calculate that, and is that the only part that I have incorrect.
 
  • #16
I think I have calculated it right, by calculating the vertical height, which gave me an answer of 1839.7, can you please tell me if this is the correct answer, or if you got something different
 
  • #17
What I did to get that is use the sin rule to get the vertical heigh (vh) which was:

vh=sin(21)*8/sin90=2.87

which i then put in the GPE formula to get:

GPE=mgh
GPE=152.04*9.8*2.87
GPE=4276.28

Which I then inserted into:
kx^2=mv^2+mgh-Fd
kx^2=608.16+4276.28-4264= 620.44

so

kx^2= 620.44
kx= 1066.97

Make both equations equal to k, so

1066.97/x=620.44/x^2

Multiply both sides by x to get

x=620.44/1066.97
x=0.58

Which I then use to calculate k by:

F=kx

k=F/x
k=1066.97/0.58
k=1839.7
 
  • #18
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What I did to get that is use the sin rule to get the vertical heigh (vh) which was:

vh=sin(21)*8/sin90=2.87

which i then put in the GPE formula to get:

GPE=mgh
GPE=152.04*9.8*2.87
GPE=4276.28
I get 4272
Which I then inserted into:
kx^2=mv^2+mgh-Fd
kx^2=608.16+4276.28-4264= 620.44
What happened to the 1/2 in the KE and stored spring energy?
 
  • #19
I get 4272

What happened to the 1/2 in the KE and stored spring energy?

I multiplied the equations by 2 two eliminate the halves
 
  • #20
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I multiplied the equations by 2 two eliminate the halves
Not on the friction and potential energy terms. You have got to be more careful about the arithmetic. That part should be a "gimme."
 
  • #21
ok, should i square root those two.

I have also tried:

0.5kx^2=0.5mv^2+mgh+Fd

0.5kx^2=304.08+4272-4264

0.5kx^2=312.08
*2

kx^2=624.16

so
kx=1066.97
kx^2=624.16

equal equations for k, rearrange fro x:

x=624.16/1066.97

x=0.58, so k=1839.6, or is this method also wrong
 
  • #22
Thank you everyone for your help, especially you Chestermiller, I could not have managed without your help, the answer was indeed 1839.6 N/m
 
  • #23
epenguin
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That equation is all right (although it should actually be an inequality).
.

Can you explain how opposing forces do not have opposite signs?
 
  • #24
PeroK
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Can you explain how opposing forces do not have opposite signs?

When the crate stops, the condition is that it doesn't rebound. At that point, you have the spring trying to push the crate back up the slope with a force of ##kx## and you have the weight of the crate and the static friction opposing this force.

If ##k## is set too high, then after being stopped, the crate will rebound back up the slope. This leads to the constraint:

##kx \le mgsin(\theta) + F_f##

To find the maximum possible value of ##k## you have equality here.
 

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