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Assume this is on Earth, so you know $g$. Don't try to use any values for $M$ or $k$. Just use symbols. Draw a free-body diagram for the system at rest. From this you can get the relationship between $M$ and $k$ and $g$. You probably also have an expression telling you the relationship between the $T$, $M$, and $k$. Do the algebra and everything should work ...

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Just a small conceptual hint will do No problem. A hint: Set up Newton's law, $\sum F=ma$. You will see that the sum of all the three forces must equal... yes, what should it equal? I'm confused with the condition at which the block will start sliding What is the difference in the equation mentioned above for a point just after it started moving, ...

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The force is the same on both springs, so each will compress by the length it would compress alone if it sees 300lb. By the way, the force is called stress, and the amount it moves is called strain.

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Using gonec's suggestion, we can solve for $k$. Now, taking the sign convention that left is negative and right is positive, lets consider the problem. In simple harmonic motion, the position is determined by a sinusoidal function, $x = x_{max}cos(\omega t + \phi)$, taking the equilibrium position to be $x=0$. Now, assuming the particle to be connected to a ...

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Actually this seems pretty simple, if you have 45N stored in the spring - this is the force that will be applied to the block when it is launched (assuming perfect perfectness perfection) from there you simply need your Second Law of motion, which states that the force is equal to the mass times the acceleration. From here you have your acceleration, at this ...

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You can actually use the conservation of energy principle to solve this.Account for the initial energy of the system...initially the potential energy of the system is 45 J + the potential energy of the block with respect to the earth and equate it to the final energy I.e. The kinetic energy and the potential energy.

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Proposed answer: the stiffness of the spring has influence on the value momentarily displayed; the stiffer, the higher. For unusually soft spring (soft enough that the scale goes down about $5$cm or more when an adult steps on it), the following analysis might allows an estimate of the stiffness of the spring. But with a normal mechanical bathroom scale, the ...

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If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...

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