A block of mass m=0.2 kg, is pushed by a spring of stiffness k=300 N/m

In summary: Nope. I'd deal directly with the energies (kinetic and potential). The only tricky bit is handling the spring action since the block can't compress past the wedge (it would slip if it tried to). So you need to find the distance from the block's initial position to the point where the spring stops pushing the block. But that's the only tricky part with this approach, and it's all in the setup.In summary, the conversation involves a student seeking help with a physics problem involving a block on a wedge with a spring, friction, and gravity. The professor provided an answer of .56m, but the student has been unable to find a solution using various formulas and online resources. Suggestions were made to approach the
  • #1
selhai97
5
0
I am currently taking a physics class and I'm studying for the final and I've been stuck on this problem and haven't been able to find help from the book nor help from online.

The professor said the answer was .56m, but I've been trying formulas for about an hour now and haven't been getting anything close to that.

Help please?


1. Homework Statement


A block of mass m=0.2 kg, is pushed by a spring of stiffness k=300 N/m, initially compressed by A=10cm, up the 30° wedge. If the friction coefficient between the block and the wedge is 0.6, how far will the block pass in m along the wedge before it stops?

Homework Equations



I think:
ω = √(k/m)
T = (2π)/ω
x = Acos(ωt + φ)

The Attempt at a Solution


[/B]
ω = √(300/.2) = 7.7
T = (2π)/7.7 = .82
x = (10)cos(7.7*.82) = 9.99

Like I said, I've been trying different formulas for about an hour now, so I don't think I'm using the right ones.

Any help would be amazing, thank you!
 
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  • #2
selhai97 said:
I am currently taking a physics class and I'm studying for the final and I've been stuck on this problem and haven't been able to find help from the book nor help from online.

The professor said the answer was .56m, but I've been trying formulas for about an hour now and haven't been getting anything close to that.

Help please?


1. Homework Statement


A block of mass m=0.2 kg, is pushed by a spring of stiffness k=300 N/m, initially compressed by A=10cm, up the 30° wedge. If the friction coefficient between the block and the wedge is 0.6, how far will the block pass in m along the wedge before it stops?

Homework Equations



I think:
ω = √(k/m)
T = (2π)/ω
x = Acos(ωt + φ)

The Attempt at a Solution


[/B]
ω = √(300/.2) = 7.7
T = (2π)/7.7 = .82
x = (10)cos(7.7*.82) = 9.99

Like I said, I've been trying different formulas for about an hour now, so I don't think I'm using the right ones.

Any help would be amazing, thank you!

The problem involves a wedge at an angle, gravity, a spring and friction? I'm not sure it's a good idea to use SHM, which would be about the frequency of oscillations of the system.
 
  • #3
PS You might want to check those numbers. Do you have a diagram?
 
  • #4
PeroK said:
PS You might want to check those numbers. Do you have a diagram?

No, no diagram, that's all the information that I was given.
 
  • #5
selhai97 said:
No, no diagram, that's all the information that I was given.

The difficulty as I see it is that on a ##30°## angle, a coefficient of friction of ##0.6## more than cancels gravity. In any case, the nett force is low. With those numbers, it's effectively just the spring, with a small variation due to the slope.

Without the slope, it should move ##20cm##.

Also, even if you take friction out of the equation, the PE of the mass is low compared to the spring. Even a mass hanging on that spring wouldn't stretch it much. You could work that out.

I'd leave it. It doesn't look right to me.
 
  • #6
PeroK said:
The difficulty as I see it is that on a ##30°## angle, a coefficient of friction of ##0.6## more than cancels gravity. In any case, the nett force is low. With those numbers, it's effectively just the spring, with a small variation due to the slope.

Without the slope, it should move ##20cm##.

Also, even if you take friction out of the equation, the PE of the mass is low compared to the spring. Even a mass hanging on that spring wouldn't stretch it much. You could work that out.

I'd leave it. It doesn't look right to me.

Alright, well thanks anyways! I really appreciate it.
 
  • #7
The problem doesn't specify if the block is attached to the spring or whether it is unattached, or what is to be considered the block's starting position (before the spring compression or after?).

Either way I'd suggest an energy approach.
 
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  • #8
gneill said:
The problem doesn't specify if the block is attached to the spring or whether it is unattached, or what is to be considered the block's starting position (before the spring compression or after?).

Either way I'd suggest an energy approach.

Yes, of course, if the mass was pushed down on the spring and not attached to it!
 
  • #9
gneill said:
The problem doesn't specify if the block is attached to the spring or whether it is unattached, or what is to be considered the block's starting position (before the spring compression or after?).

Either way I'd suggest an energy approach.

You wouldn't use something like this right?
max = -mgsinθ
vxf2 = vxi2 + 2aΔx
 
  • #10
selhai97 said:
You wouldn't use something like this right?
max = -mgsinθ
vxf2 = vxi2 + 2aΔx
Nope. I'd deal directly with the energies (kinetic and potential). The only tricky bit is handling the spring action since the block and spring become disengaged at some point.
 
  • #11
gneill said:
Nope. I'd deal directly with the energies (kinetic and potential). The only tricky bit is handling the spring action since the block and spring become disengaged at some point.

So PE = mgh and KE = (½)mv2?

Are there other equations that involve the A and k and μ?

I'm sorry if I'm so lost, physics is my hardest subject.
 
  • #12
selhai97 said:
So PE = mgh and KE = (½)mv2?

Are there other equations that involve the A and k and μ?

I'm sorry if I'm so lost, physics is my hardest subject.
Look into formulas pertaining to the potential energy stored by a spring and the energy lost to friction.
 

FAQ: A block of mass m=0.2 kg, is pushed by a spring of stiffness k=300 N/m

1. How do you calculate the force exerted by the spring on the block?

The force exerted by the spring on the block can be calculated using the formula F = kx, where k is the stiffness of the spring and x is the displacement of the block from its equilibrium position.

2. What is the equilibrium position of the block when it is pushed by the spring?

The equilibrium position of the block is the position where the spring force and the force of gravity acting on the block are balanced, resulting in no net force on the block.

3. How does the displacement of the block affect the force exerted by the spring?

The force exerted by the spring is directly proportional to the displacement of the block from its equilibrium position. This means that the greater the displacement, the greater the force exerted by the spring.

4. Can the stiffness of the spring affect the motion of the block?

Yes, the stiffness of the spring can affect the motion of the block. A stiffer spring will exert a greater force on the block, resulting in a faster acceleration and a larger displacement.

5. What is the relationship between the mass of the block and the force exerted by the spring?

The mass of the block does not directly affect the force exerted by the spring. However, it does affect the resulting acceleration of the block. A heavier block will require a greater force from the spring to achieve the same acceleration as a lighter block.

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