Newton Laws - Finding a constant

AI Thread Summary
The discussion revolves around calculating the stiffness constant k of a material when a metallic cone penetrates it. The problem specifies that the cone, weighing 0.3 kg, falls freely and penetrates 5 cm into the material, with the force exerted by the material described by kx². Initial calculations suggested k should be 24696 N/m², but discrepancies arose due to integration limits and frame of reference issues. Ultimately, the correct value for k was confirmed as 24696 N/m², emphasizing that the final answer remains consistent regardless of the chosen frame of reference. The conversation highlights the importance of careful consideration of forces and integration limits in physics problems.
bluewood
Messages
9
Reaction score
0

Homework Statement


To find the stiffness of a material, a block of that material is placed 30 cm under a metallic cone with 0.3 kg; the cone is free falling from rest, penetrating a distance x in the block until it stops. It is known that when the cone penetrates in the block the force of the block exerced on the cone is kx2 where k is a constant that depends on the stiffness to penetration of the material; if the cone penetrates a distance x = 5 cm, find the value of the constant k.

An added restriction that I'm adding to the problem is to not use the relations between Work and Energy (at least directly) -> that subject wasn't approached yet.


Homework Equations


The derivatives of x(t) and v(t), and:
a = v\frac{dv}{dx}
F = m a


The Attempt at a Solution


According to the solutions and to the Work-Energy relations, k should be 24696 N/m2. But I've tried several times and the values don't match:
F - W = m a
k {x}^{2} - 9.8 m = m a
k {x}^{2} - m 9.8 = m v \frac{dv}{dx}
\int_{0.05+0.3}^{0}k {x}^{2} - m 9.8\,dx = \frac{1}{2} ({0}^{2}-{0}^{2})
-\frac{343 k - 24696}{24000} = 0
k = 72
 

Attachments

  • falling.jpg
    falling.jpg
    4.5 KB · Views: 429
Physics news on Phys.org
Check your limits of integration. You cannot put both forces under the same integral. Gravity acts over the entire distance of 0.35 cm, but the "stiffness" force acts over 0.05 cm only.
 
Thanks for the remark, I've managed to find the correct solution. I've tried several times before, and one of the steps that was confusing me before was when I was calculating the speed of the cone before hitting the cone.

If I considered a frame of reference where g = -9.8 m/s2, the speed before hitting the block would be an imaginary value, but using a different frame where g = +9.8, the speed would be a real value - ignoring those facts the numeric value is the same. It is kinda awkward, since using the first referential most of the people would think their calculations were wrong (EDIT: see the end of this post).

The other thing was the net force acting on the cone, wether the normal reaction existed or not. Anyway, here's the calculations in case anyone wants the solution (I'm considering "up" as positive and x = 0 when the cone is inside the block):

1. Finding the speed of the cone before hitting the block

v\frac{dv}{dx} = -9.8

\int_{0}^{v_f}v\,dv = \int_{x+0.30}^{x}-9.8\,dx

\left[ \frac{v^2}{2} \right]_{0}^{v_f} = \left[-9.8 x \right]_{x+0.30}^{x}

{v_f}^2 = 2 \times -9.8 \times (x - (x + 0.30))

v_f = \pm \sqrt{5.88}

For this case we're interested in the negative value.


2. Finding k

F - W = m a

kx^2 - 9.8 m = m a

\int_{0}^{0.05}(kx^2 - 9.8 m)\,dx = \int_{0}^{- \sqrt{5.88}} mv\,dv

\left[ \frac{kx^3}{3} - 9.8mx \right]_{0}^{0.05} = \left[ \frac{mv^2}{2} \right]_{0}^{-5.88}

\frac{k}{24000} - 0.49m = 2.94 m

k = 82320 m

Since m = 0.3 kg:

k = 24696 N/m^2

EDIT: Corrected a small mistake in the right integral of step 1. The speed value is always real no matter what frame of reference is chosen.
 
Last edited:
bluewood said:
If I considered a referential where g = -9.8 m/s2, the speed before hitting the block would be an imaginary value, but using a different referential where g = +9.8, the speed would be a real value - ignoring those facts the numeric value is the same. It is kinda awkward, since using the first referential most of the people would think their calculations were wrong.
The correct answer should be independent of your frame of reference. In other words, the block will hit the material with the correct speed regardless of whether you (or anyone else) choose "up" as positive or "down" as positive.

The kinematic equation to use is

2 a Δx = v2 - v02

with v0 = 0, we get
v=\sqrt{2a\Delta x}

Note that both the displacement and the acceleration vectors point "down". If we choose "down" as positive, then both quantities under the radical are positive and there is no problem. If we choose "up" as positive, then both quantities under the radical are negative, however their product is a positive number and there still is no problem. :wink:
 
You are correct. Actually, that imaginary value in my previous calculations in the speed was due to an error in the limits of integration. I've tried several (fixed) frames of reference and all yielded the correct value for the speed.

Thanks :biggrin:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top