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Force created from kinetic energy

  • Thread starter pantsik
  • Start date
5
0
I study physics and I cannot find the following themes (they are quite similar) anywhere in physics textbooks:

Problem 1: I drive my car (without a seatbelt) with a velocity v and my body mass is m.
Suddenly I collide in a unmovable wall. Consider my car to be completely stiff. Due to the conservation of momentum my body will continue to move forward with a velocity v and I will have a kinetic energy of K = 1/2 mv^2.
On the wall is installed vertically a well adjusted scale.
What is the indication of the scale when my body eventually stops on it (the scale) relatively to my body weight (B = mg) ?
If I apply the conservation of energy theorem: 1/2 mv^2 = 1/2 kx^2 (where k is the scale's constant and x the spring's movement) I find a result which depends on k or x which doesn't seem right to me [F = (v / sqrt(xg)) * B where B is my gravitational force and F is the force created from the collision]. The reason is that every scale should show the same indication regardless of its k or x.
Can anyone give me the right equation ?

Problem 2: I jump from h meters above the ground to the previous scale. What would be the scale's indication (the higher one) relatively to my weight ?

Forgive my poor english.
 

Answers and Replies

193
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F=MA

If A is gravity then the F imparted on the scale is calibrated to indicate M.

If you jump then the scales slow you down so the F imparted on the scale is Gravity + Your deceleration from terminal velocity to stationary, jumping straight legged will result in a higher indication than jumping bent legged where some deceleration is in your legs rather than the scales.

In the unmoving objects situation then the deceleration is infinite as is the force.

Try weighing yourself in a lift. Your indicated weight will change according to total acceleration (gravity + Lift Acceleration)
 
5
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F=MA

If A is gravity then the F imparted on the scale is calibrated to indicate M.

If you jump then the scales slow you down so the F imparted on the scale is Gravity + Your deceleration from terminal velocity to stationary, jumping straight legged will result in a higher indication than jumping bent legged where some deceleration is in your legs rather than the scales.

In the unmoving objects situation then the deceleration is infinite as is the force.

Try weighing yourself in a lift. Your indicated weight will change according to total acceleration (gravity + Lift Acceleration)
Thanks for your reply, but it is too general. Can you give me the equations?
 
400
1
A spring scale measures force, and in both examples the impact force is F = dP/dt = mdv/dt = ma. You know or can calculate the velocity at impact in both cases, so you know dP. Does dt depend on the spring constant?
 
5
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A spring scale measures force, and in both examples the impact force is F = dP/dt = mdv/dt = ma. You know or can calculate the velocity at impact in both cases, so you know dP. Does dt depend on the spring constant?
In the case of horizontal movement, dv/dt = 0 (the velocity is constant).
In the case of vertical movement, dv/dt = g, meaning F = mg.
In both cases you don't find out the force produced by the kinetic energy.
 
5
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In the case of horizontal movement, dv/dt = 0 (the velocity is constant).
In the case of vertical movement, dv/dt = g, meaning F = mg.
In both cases you don't find out the force produced by the kinetic energy.
If you mean the force created on the spring, then you 're right. F = ma where "a" is the deceleration of the mass m.
But what about the equations?
 
400
1
You already wrote down the main equation, mv^2/2 = kx^2/2. That will give you the maximum distance X for an incident speed V. The spring force at that compresssion is kX, and in units of body weight it's kX/mg.

If you think the answer doesn't depend on k or X, consider the case of an extremely stiff spring like a brick wall. Does your body feel more decelleration force in that case, compared to the case of a soft spring? Also consider the purpose of suspension springs on cars - does an extremely stiff spring (no suspension) give a harsher ride than a soft spring?
 
5
0
You already wrote down the main equation, mv^2/2 = kx^2/2. That will give you the maximum distance X for an incident speed V. The spring force at that compresssion is kX, and in units of body weight it's kX/mg.

If you think the answer doesn't depend on k or X, consider the case of an extremely stiff spring like a brick wall. Does your body feel more decelleration force in that case, compared to the case of a soft spring? Also consider the purpose of suspension springs on cars - does an extremely stiff spring (no suspension) give a harsher ride than a soft spring?
You are right. So the answer to my question is that every scale measures a different force when the measuring is made off-balance.
 

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