Question

a big lump of meat of mass 5kg is hung from a spring balance in an elevator, find the reading of the balance if; (I) the elevator is moving with a steady speed. (ii)the elevator is moving upwards with acceleration of 0.2m/s
(g= 10 m/s^2)​

Answer 1

Given:

A big lump of meat of mass 5kg is hung from a spring balance in an elevator.

To find:

Reading of the balance if;

• the elevator is moving with a steady speed.
• the elevator is moving upwards with acceleration of 0.2m/s².

Calculation:

1st case:

• When the lift is moving up/down with a steady speed, the mass will not experience any pseudo-force.

• Hence, the reading will be equal to weight of the mass.

$\rm \therefore \: reading = mg=5g = 50N$

So, reading will be 50N.

2nd case:

• When elevator is moving upwards with 0.2 m/s² , it will experience an pseudo-force downwards.

$\rm\therefore \: reading = mg + ma$

$\rm \implies\: reading = m(g + a)$

$\rm \implies\: reading = 5(10+ 0.2)$

$\rm \implies\: reading = 5 \times 10.2$

$\rm \implies\: reading = 51 \: N$

So, reading will be 51 N.

Answer 2

Answer:

Given:

A big lump of meat of mass 5kg is hung from a spring balance in an elevator.

To find:

Reading of the balance if;

the elevator is moving with a steady speed.

the elevator is moving upwards with acceleration of 0.2m/s².

Calculation:

1st case:

When the lift is moving up/down with a steady speed, the mass will not experience any pseudo-force.

Hence, the reading will be equal to weight of the mass.

reading=mg=5g=50N

So, reading will be 50N.

2nd case:

When elevator is moving upwards with 0.2 m/s² , it will experience an pseudo-force downwards.

\rm\therefore \: reading = mg + ma∴reading=mg+ma

\rm \implies\: reading = m(g + a)⟹reading=m(g+a)

\rm \implies\: reading = 5(10+ 0.2)⟹reading=5(10+0.2)

\rm \implies\: reading = 5 \times 10.2⟹reading=5×10.2

So, reading will be 51 N.