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Potential energy lets us do work in the present to change things in the future. If energy is currency, then potential energy is money in the bank.

A grandfather clock works by a pendulum system and keeps time because the pendulum has a constant frequency. The grandfather clock in my room has stopped, but I'm too lazy to get out of bed to start the pendulum oscillating again. Hence I throw a sticky piece of gum at the pendulum to get it moving again. My piece of gum has a mass of 10 g, hits the pendulum inelastically with a speed of 10 m/s, and sticks there. Admittedly disgusting. But, more importantly, if the pendulum has a mass of 90 g, how high does the pendulum on my grandfather clock go **in meters**?

**Details and assumptions**

- The acceleration due to gravity is \(9.8~m/s^2\).
- Treat the grandfather clock as a simple pendulum.

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An object of mass \(m=9\text{ kg}\) is released from rest at point \(A,\) and slides down a long, frictionless, \(h=60\text{ m}\) high slide. Then it enters the horizontal surface from point \(B\) to \(C,\) which is not frictionless, and comes to a complete stop at point \(C.\) If the coefficient of kinetic friction between the object and the surface is \(\mu=0.1,\) what is the horizontal distance \(x\) between points \(B\) and \(C?\)

The gravitational acceleration is \(g=10\text{ m/s}^2.\)

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An object of mass \(4\text{ kg}\) initially at rest falls freely towards a huge spring that is \(100\text{ m}\) below. If the spring constant is \(k=16\text{ N/m},\) what is the maximum change in the spring's length?

Air resistance is negligible and gravitational acceleration is \(g=10\text{ m/s}^2.\)

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A meteor of mass \(m=800\text{ kg}\) enters the earth's atmosphere with a velocity of \(500\text{ m/s},\) and falls vertically toward the ground. The meteor burns while falling, and its mass decreases at a rate of \(5\text{ kg/s}.\) If the meteor falls with a constant velocity, how much gravitational potential energy will it lose during \(16\) seconds?

**Note**

- The height of the meteor when it enters the atmosphere is \(1000\text{ km},\)
- The gravitational acceleration is \(g=10\text{ m/s}^2.\)

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