QuestionJune 16, 2025

30. A12 -kg projectile is launched with an initial vertical speed of 20m/s It rises to a maximum height of 18 m above the launch point. What is the change in mechanical energy caused by the dissipative (air) resistive force on the projectile during this ascent?

30. A12 -kg projectile is launched with an initial vertical speed of 20m/s It rises to a maximum height of 18 m above the launch point. What is the change in mechanical energy caused by the dissipative (air) resistive force on the projectile during this ascent?
30. A12 -kg projectile is launched with an initial vertical speed of 20m/s It rises to a maximum height of 18 m above the launch point. What is the change in mechanical energy caused by the dissipative (air) resistive force on the projectile during this ascent?

Solution
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Answer

The change in mechanical energy is -283.2\, \text{J}. Explanation 1. Calculate Initial Mechanical Energy Initial kinetic energy KE_i = \frac{1}{2}mv^2 = \frac{1}{2}(12\, \text{kg})(20\, \text{m/s})^2 = 2400\, \text{J}. 2. Calculate Final Mechanical Energy At maximum height, velocity is zero. Potential energy PE_f = mgh = (12\, \text{kg})(9.8\, \text{m/s}^2)(18\, \text{m}) = 2116.8\, \text{J}. 3. Determine Change in Mechanical Energy Change in mechanical energy \Delta E = PE_f - KE_i = 2116.8\, \text{J} - 2400\, \text{J} = -283.2\, \text{J}.

Explanation

1. Calculate Initial Mechanical Energy<br /> Initial kinetic energy $KE_i = \frac{1}{2}mv^2 = \frac{1}{2}(12\, \text{kg})(20\, \text{m/s})^2 = 2400\, \text{J}$.<br /><br />2. Calculate Final Mechanical Energy<br /> At maximum height, velocity is zero. Potential energy $PE_f = mgh = (12\, \text{kg})(9.8\, \text{m/s}^2)(18\, \text{m}) = 2116.8\, \text{J}$.<br /><br />3. Determine Change in Mechanical Energy<br /> Change in mechanical energy $\Delta E = PE_f - KE_i = 2116.8\, \text{J} - 2400\, \text{J} = -283.2\, \text{J}$.
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