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Chapter 5: Problem 60

5.60 A man throws a rock of mass \(m=0.325 \mathrm{~kg}\) straight up into the air. In this process, his arm does a total amount of work \(W_{\text {net }}=115 \mathrm{~J}\) on the rock. Calculate the maximum distance, \(h\), above the man's throwing hand that the rock will travel.

Short Answer

Expert verified
Answer: The maximum height the rock will reach above the man's hand is approximately 3.6 meters.

Step by step solution

01

Calculate the initial kinetic energy of the rock

To find the maximum height, we first need the initial kinetic energy of the rock. We know the net work done on the rock, \(W_{\text{net}}\), and can use the work-energy principle to find the initial kinetic energy, \(K_1\). The principle states that the work done on an object is equal to the change in kinetic energy, therefore, \(W_{\text {net}}=K_1 - K_0\). Since the rock is initially at rest, \(K_0 = 0\). So, the initial kinetic energy is equal to the net work done: \(K_1 = W_{\text {net}} = 115\ \mathrm{J}\).
02

Calculate the maximum potential energy of the rock

At the maximum height, the rock will momentarily come to rest before falling back down, which means its final kinetic energy, \(K_2\), will be zero. Therefore, all of its initial kinetic energy will convert into potential energy, \(U_2\), due to gravity. Using the work-energy principle again, \(K_1 - K_2 = U_2 - U_1\). Since the rock is initially at the man's hand height, we can consider its initial potential energy, \(U_1\), to be zero. Thus, the maximum potential energy is equal to the initial kinetic energy: \(U_2 = K_1 = 115\ \mathrm{J}\).
03

Calculate the maximum height of the rock above the man's hand

Now, we can use the formula for gravitational potential energy, \(U=mgh\), to find the maximum height, \(h\). At the maximum height, the potential energy will be equal to the initial kinetic energy, so \(U_2 = mgh\). Rearranging this equation, we get: \(h = \frac{U_2}{mg}\) Substitute the given values for \(m\), \(U_2\), and \(g=9.8\ \mathrm{m/s^2}\) (the acceleration due to gravity): \(h = \frac{115\ \mathrm{J}}{0.325\ \mathrm{kg} * 9.8\ \mathrm{m/s^2}}\)
04

Calculate the maximum height

Finally, perform the calculation: \(h \approx 3.6\ \mathrm{m}\) Therefore, the maximum distance above the man's throwing hand that the rock will travel is approximately 3.6 meters.

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