Question

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 1

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

                                  $a=\frac{(T -Mg)}{M}\\ = \frac{(8300 -7350)}{750 } \\=1.267m/s^2$

2.The normal reaction of the man is

                               $R =mg+ma \\ =(75)\times(9.8)+(75)\times(1.267) \\ =735+95 \\ =830N$

3.so The upward velocity of the elevator is given by

                               $v =u +at \\= 0 +(1.267m/s^2)\times(3s) \\ =3.801m/s$