Question

A block of mass m kg is kept on a weighing machine in an elevator. If the elevator is retarding upward by a m/s2
the reading of weighing machine is​

Answer 1

ANSWER:

Reading of the weighing machine = $\frac{m(a+g)}{g}$

EXPLANATION:

Let us assume that the lift or elevator is moving up with an acceleration of “a” m/$s^{2}$

A weight of mass “m” is kept on a weighing machine, in the elevator

The reactionary force acting on the body is given by

R = ma + mg

Let us take the mass common we get:

R = m(a+g)

Therefore, the reactionary force is given by R = m(a+g)

Now as for the weight reading of the object when the elevator goes up is

Reading = $\frac{m(a+g)}{g}$

Answer 2

Answer:

THE READING OF THE WEIGHING MACHINE IS  $\frac{m(g+a)}{g}$

Explanation:

Weighing machine reads the mass of the body if the body is not under the influence of any other force but greater than the mass if it is under a force as given in this case

Using the NEWTONS LAWS OF MOTION

If the net acceleration of the block of mass m is considered as A then

mA = mg - m(-a) as acceleration of the elevator is in the direction opposite to the conventional direction

This gives

A = g+a

A weighing machine reads the value depending on the acceleration in the lift hence it reads the value $\frac{m(g+a)}{g}$ where the equation can be analysed as effective mass = $\frac{force  experienced}{acceleration due to gravity}$

HENCE THE ANSWER IS $\frac{m(g+a)}{g}$