1000.kg ^^ (22.6 $\mu_s$=0.2 $\mu_k$=0.1 96m R Find safe v for the above area. \begin{cases} 1) lowest safe speed\\ 2) highest safe speed \end{cases} (entire range) 22.6
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5 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity. Given: m = 1000 kg KE = 0.1 Mk = 0.1 * 96m * 22.6 PM = 0.1 * 96 * 22.6 = 217.44 PM Now, plug in the values and solve for v: 217.44 PM = 0.5 * 1000 * v^2 217.44 PM = 500v^2 v^2 = Show more…
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