Questions: The energy of an object is given by E=mg h + 1/2 m(dh/dt)^2 where m is the (constant) mass of the object, g is constant, and h is the height of the object. If the energy of the object is increasing at a rate of 50 ft. lbs. / sec, find the acceleration of the object the instant its velocity is 10 ft / sec.

The energy of an object is given by E=mg h + 1/2 m(dh/dt)^2 where m is the (constant) mass of the object, g is constant, and h is the height of the object. If the energy of the object is increasing at a rate of 50 ft. lbs. / sec, find the acceleration of the object the instant its velocity is 10 ft / sec.
Transcript text: (b) The energy of an object is given by $E=m g h+\frac{1}{2} m\left(\frac{d h}{d t}\right)^{2}$ where $m$ is the (constant) mass of the object, $g$ is constant, and $h$ is the height of the object. If the energy of the object is increasing at a rate of $50 \mathrm{ft} . \mathrm{lbs} . / \mathrm{sec}$, find the accelaration of the object the instant its velocity is $10 \mathrm{ft} / \mathrm{sec}$.
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Solution

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Solution Steps

Step 1: Differentiate the energy equation with respect to time

Given the energy equation E = mgh + (1/2)m(dh/dt)^2, we differentiate it with respect to time (t):

dE/dt = mg(dh/dt) + m(dh/dt)(d²h/dt²)

Step 2: Substitute the given values

We are given dE/dt = 50 ft.lbs./sec and dh/dt = 10 ft/sec. We need to find d²h/dt², which represents acceleration. Substituting the given values:

50 = mg(10) + m(10)(d²h/dt²)

Step 3: Simplify and solve for acceleration

We can simplify the equation by dividing both sides by 10m (since mass 'm' is constant and non-zero):

5/m = g + d²h/dt²

Since the problem asks for acceleration at a specific velocity and rate of energy increase, but doesn't provide values for _m_ or _g_, we express acceleration in terms of those variables:

d²h/dt² = 5/m - g

Final Answer:

5/m - g ft/sec²

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