A woman weighing 75 kg stands on a spring scale in elevator. During the first 3 seconds of motion from rest, the tension T in the hoisting cable is 8300 N. The total mass of the elevator , woman scale is 750 kg.
1.Find the vertical acceleration of the elevator.
2.Find the reading R of the scale in Newtons during this interval.
3.Find the upward velocity of the elevator at the end of a 3 seconds.
Answer
Given that
The mass of the man (m) =75kg
The total mass of the (elevator, man, and scale is) (M) =750kg.
The tension T is (T) =8300 N
Acceleration due to gravity
g =9.8m/s2
The time period t =3s
The initial velocity of the elevator
u =0 m/s
The elevator is moving in the upward direction means it is accelerating in the upward direction then
The tension is (T) = Mg+Ma
1.Then the acceleration of the elevator is
"a=\\frac{(T -Mg)}{M}\\\\\n\n = \\frac{(8300 -7350)}{750\n}\n \\\\=1.267m\/s^2"
2.The normal reaction of the man is
"R =mg+ma\n\n \\\\ =(75)\\times(9.8)+(75)\\times(1.267)\n\n \\\\ =735+95\n\n \\\\ =830N"
3.so The upward velocity of the elevator is given by
"v =u +at\n\n \\\\= 0 +(1.267m\/s^2)\\times(3s)\n\n \\\\ =3.801m\/s"
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