An elevator and its load weigh 160N.?

find the tension in the supporting cable,if after it was moving downward at 30m/s,it's brought to rest with const.acc.in a distance of 20m.With what force will the feet of the passenger press downward on the elevator floor,when the elevator has an acc.of 30m/s2 upward,if the passenger weighs,a.)16N;b.)640N;c.)64N?

An elevator and its load weigh 160N.?
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.





For each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.





16N/(9.8 m/s^2)=1.63 kg


1.63 kg*30 m/s^2=(force of feet)-16N


Force of feet=64.9N





Answers for the second and third scenario:


b) 2600N


c) 260N
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