Question

if cabin with descending vertically with acceleration and action by mass and on floor of is 9mg by 10 then a equal to​

Answer 1

CORRECT QUESTION:

If cabin with descending vertically with acceleration and normal reaction on floor 9mg/10 then value of 'a' equal to?

Calculation:

• First see the Free-Body Diagram of the lift as shown in image.

Let acceleration be 'a' ,

• Since the lift is descending vertically downwards, it will experience a pseudo-force upwards.

$N = mg - ma$

$\implies \dfrac{9mg}{10} = mg - ma$

$\implies ma= mg - \dfrac{9mg}{10}$

$\implies ma= \dfrac{(10 - 9)mg}{10}$

$\implies ma= \dfrac{mg}{10}$

$\implies a= \dfrac{g}{10}$

So, value of acceleration is g/10 .

Answer 2

Answer:

$CORRECT QUESTION:</p><p></p><p>If cabin with descending vertically with acceleration and normal reaction on floor 9mg/10 then value of 'a' equal to?</p><p></p><p>Calculation:</p><p></p><p>First see the Free-Body Diagram of the lift as shown in image.</p><p></p><p>Let acceleration be 'a' ,</p><p></p><p>Since the lift is descending vertically downwards, it will experience a pseudo-force upwards.</p><p></p><p>N = mg - maN=mg−ma</p><p></p><p>\implies \dfrac{9mg}{10} = mg - ma⟹109mg=mg−ma</p><p></p><p>\implies ma= mg - \dfrac{9mg}{10}⟹ma=mg−109mg</p><p></p><p>\implies ma= \dfrac{(10 - 9)mg}{10}⟹ma=10(10−9)mg</p><p></p><p>\implies ma= \dfrac{mg}{10}⟹ma=10mg</p><p></p><p>\implies a= \dfrac{g}{10}⟹a=10g</p><p></p><p>So, value of acceleration is g/10 .</p><p></p><p>$