Question
An elevator and its load have a combined mass of $1600 \mathrm{~kg}$. Find the pull (or tension) force supplied by in the supporting cable when the elevator, originally moving downward at $12 \mathrm{~m} / \mathrm{s}$, is brought to rest with constant acceleration in a distance of $42 \mathrm{~m}$.
Step 1
We can use the equation of motion to find this: \[a_y = \frac{v_{f_y}^2 - v_{i_y}^2}{2d_y}\] where \(a_y\) is the acceleration, \(v_{f_y}\) is the final velocity, \(v_{i_y}\) is the initial velocity, and \(d_y\) is the displacement. Show more…
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An elevator and its load have a combined mass of $1600 \mathrm{~kg}$. Find the tension in the supporting cable when the elevator, originally moving downward at $12.0 \mathrm{~m} / \mathrm{s}$, is brought to rest with constant acceleration in a distance of $42.0 \mathrm{~m}$.
An 1600 kg elevator was moving downward at 12 m/s, and brought to rest with constant acceleration in a distance of 42 m. Find the tension in the cable supporting the elevator during this slow down.
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