Part A Select the correct equations that show that a 2500000 -J change in kinetic energy occurs for an airplane that is moved 500 m in takeoff by a sustained force of 5000 N . W \( =F^{2} / g=(5000 \mathrm{~N})^{2} /(10 \mathrm{~N} / \mathrm{m})=2500000 \mathrm{~J} \) \( W=F d / g=(5000 \mathrm{~N})(500 \mathrm{~m}) /(10 \mathrm{~N} / \mathrm{m})=2500000 \mathrm{~J} \) \( W=F d=(5000 \mathrm{~N})(500 \mathrm{~m})=2500000 \mathrm{~J} \) \( W=d^{2} g=(500 \mathrm{~m})^{2}(10 \mathrm{~N} / \mathrm{m})=2500000 \mathrm{~J} \) Submit Request Answer
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The Deep Dive
To find the work done on the airplane during takeoff, we can use the equation \( W = F \cdot d \), where \( W \) is work, \( F \) is the applied force, and \( d \) is the distance moved. With a force of 5000 N and moving 500 m, we calculate \( W = (5000 \, \text{N})(500 \, \text{m}) = 2500000 \, \text{J} \). So, the correct equation showing this energy change is \( W = F \cdot d = 2500000 \, \text{J} \). The other formula options include unnecessary divisions and other configurations, which may confuse the kinetic energy calculation. It's essential to remember that work is simply the force multiplied by the distance moved in the direction of the force, which gives clear and correct insight into the physics involved during the airplane's takeoff.