An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments. The elevator accelerates upwards from rest to a speed of $$\(2 \mathrm{~m} \mathrm{~s}^{-1}\)$$ over a period of $$\(1.5 \mathrm{~s}\)$$ and then travels at this speed for $$\(4.5 \mathrm{~s}\)$$, before decelerating to rest over a period of $$\(1 \mathrm{~s}\)$$. The elevator then remains at rest for $$\(6 \mathrm{~s}\)$$, before accelerating to a speed of $$\(V \mathrm{~m} \mathrm{~s}^{-1}\)$$ downwards over a period of $$\(2 \mathrm{~s}\)$$. The elevator travels at this speed for a period of $$\(5 \mathrm{~s}\)$$, before decelerating to rest over a period of $$\(1.5 \mathrm{~s}\)$$ The combined weight of the elevator and passengers on its upward journey is $$\(1500 \mathrm{~kg}\)$$. Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_m21_qp_42 Year：2021 Question No：4(c)
Answer:
Acceleration \(=-2 \mathrm{~m} \mathrm{~s}^{-2}\)
\(T-1500 g=1500 \times(-2)\)
\(T=12000 \mathrm{~N}\)
\(T-1500 g=1500 \times(-2)\)
\(T=12000 \mathrm{~N}\)
Knowledge points:
4.2.2.3 the gradient of a velocity–time graph represents acceleration
4.4.1 apply Newton’s laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
4.4.3 solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
Solution:
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