Question

an engineer is trying to determine the height from which she needs to release a 55kg wrecking ball such that the wrecking ball will be traveling 10.7m/s when it strikes the building at the position shown. neglect drag. list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. include the earth as part of the system. initial: unit work? unit loss? unit final: unit from what height above the ground does the engineer need to release the ball? m

Explanation:

Step1: Identify Energy Types

The system includes the ball and Earth, so initial energy is gravitational potential energy (\(U_{g,i}\)) and kinetic energy (\(K_i\)). Since it's released from rest, \(K_i = 0\). Initial potential energy: \(U_{g,i}=mgh_1\), where \(h_1\) is the initial height above the reference (ground at the building's strike point? Wait, the final height is 5.7m? Wait, no—wait, the problem is to find the release height \(h\) such that when it strikes the building (at 5.7m above ground?), the speed is 10.7m/s. Wait, let's use conservation of energy. The initial state: release from height \(h\) (above ground), so initial height relative to the strike point is \(h - 5.7\)m? Wait, no—maybe the reference is the ground. Wait, let's define: let the final position (strike point) be at height \(y_f = 5.7\)m, and initial position \(y_i = h\) (what we need to find). Initial energy: \(K_i + U_{g,i}=0 + mgh\). Final energy: \(K_f + U_{g,f}=\frac{1}{2}mv_f^2 + mgy_f\). Since no non - conservative work (neglect drag, and tension does no work as it's perpendicular to displacement), work \(W = 0\), loss \(L = 0\).

Step2: Apply Conservation of Energy

Conservation of energy: \(K_i+U_{g,i}+W = K_f+U_{g,f}+L\). Since \(K_i = 0\), \(W = 0\), \(L = 0\), so \(mgh=\frac{1}{2}mv_f^2 + mgy_f\). We can cancel \(m\): \(gh=\frac{1}{2}v_f^2+gy_f\). Solve for \(h\): \(h=\frac{v_f^2}{2g}+y_f\). \(g = 9.8m/s^2\), \(v_f = 10.7m/s\), \(y_f = 5.7m\).

Step3: Calculate \(h\)

First, calculate \(\frac{v_f^2}{2g}=\frac{(10.7)^2}{2\times9.8}=\frac{114.49}{19.6}\approx5.84m\). Then \(h = 5.84+5.7 = 11.54m\)? Wait, wait—maybe the initial height is relative to the final position. Wait, maybe the reference is the final position (strike point) as \(y = 0\). Then initial height is \(h\) (above strike point), final height \(y_f = 0\), initial energy \(U_{g,i}=mgh\), \(K_i = 0\); final energy \(K_f=\frac{1}{2}mv^2\), \(U_{g,f}=0\). Then \(mgh=\frac{1}{2}mv^2\), so \(h=\frac{v^2}{2g}\). Then total height above ground is \(h + 5.7\). Wait, that makes more sense! Because the strike point is at 5.7m above ground. So initial height relative to strike point: \(h_{rel}=\frac{v^2}{2g}\), then total height \(H=h_{rel}+5.7\).

Calculate \(h_{rel}=\frac{(10.7)^2}{2\times9.8}=\frac{114.49}{19.6}\approx5.84m\). Then \(H = 5.84+5.7 = 11.54m\approx11.5m\) (or more accurately, let's recalculate \(\frac{10.7^2}{2\times9.8}\): \(10.7\times10.7 = 114.49\), \(2\times9.8 = 19.6\), \(114.49\div19.6 = 5.8413\). Then \(5.8413+5.7 = 11.5413\approx11.5m\) or 11.54m.

Wait, let's check the energy types:

Initial: Gravitational Potential Energy (relative to ground) \(U_{g,i}=mgh\) (h is what we find, mass \(m = 55kg\), but when we cancel m, it's not needed for the height calculation). Work: 0 (no non - conservative forces), Loss: 0 (neglect drag). Final: Kinetic Energy \(K_f=\frac{1}{2}mv_f^2\) and Gravitational Potential Energy \(U_{g,f}=mg\times5.7\) (since final height is 5.7m above ground).

But for the height calculation, using \(mgh=\frac{1}{2}mv^2+mg\times5.7\), divide by m: \(gh=\frac{v^2}{2}+g\times5.7\), so \(h=\frac{v^2}{2g}+5.7\). Plugging \(v = 10.7m/s\), \(g = 9.8m/s^2\):

\(\frac{(10.7)^2}{2\times9.8}=\frac{114.49}{19.6}\approx5.84\)

\(h=5.84 + 5.7=11.54m\approx11.5m\) (or more precisely 11.54m)

Answer:

\(11.5\) (or \(11.54\))