Question
$\mathrm{A} 550 \mathrm{kg}$ elevator accelerates upward at $1.2 \mathrm{m} / \mathrm{s}^{2}$ for the first $15 \mathrm{m}$ of its motion. How much work is done during this part of its motion by the cable that lifts the elevator?
Step 1
We can use Newton's second law, which states that force is equal to mass times acceleration: $F = m \cdot a$ $F = 550 \, \mathrm{kg} \cdot 1.2 \, \mathrm{m/s^2}$ $F = 660 \, \mathrm{N}$ Show more…
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A 550 kg elevator accelerates upward at 1.2 m/s2 for the first 15 m of its motion. How much work is done during this part of its motion by the cable that lifts the elevator?
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