An elevator of weight W is initially at rest on the first floor of a building. I

Question

An elevator of weight W is initially at rest on the first floor of a building. It moves upward and passes the second floor with speed v0. It continues upwards and finally stops at the fourth floor. The distance between adjacent floors is H.

Write an expression for the work done on the elevator by the cable over the entire trip in terms of the given quantities only. Indicate whether this work is positive or negative.

Does your result imply that the force on the elevator by the cable is always equal in magnitue to the weight of the elevator? explain

Explanation / Answer

work done by gravity on elevartor is negative as the displacement is in opposite direction as that of gravity force on elevator by cable is not always equal to mg it is m(g+a) is elevator is accelerated upwards W=mg*(3H)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.