Question

Please find the reading of the spring balance.

Answer 1

$\bigstar\huge\boxed{\mathtt{\red{Solution:}}}$

As per the given question ,

• Reading of the spring balance = 50 kg

$\bigstar\large\boxed{\mathtt{\pink{Case I:}}}$

As the lift is moving in upward direction the normal force  acts in upward direction the weight of the person acts in downward direction  

hence ,

$\rightarrow\mathtt{N - mg= ma}$

$\rightarrow\mathtt{N =mg + ma}$

$\rightarrow\mathtt{N =m(g + a)}$

As the lift is moving with constant velocity the acceleration of the elevator is zero

$\rightarrow\mathtt{N =mg}$

$\rightarrow\mathtt{\orange{N =50 N}}$

The reading of the spring balance when lift is moving with uniform velocity in upward direction is 50 N

____________________

$\bigstar\large\boxed{\mathtt{\pink{Case II:}}}$

Let us consider that the lift is moving in upward direction with constant acceleration 'a'

hence ,

$\rightarrow\mathtt{N - mg= ma}$

$\rightarrow\mathtt{N =mg + ma}$

$\rightarrow\mathtt{N =m(g + a)}...(1)$

we know that ,

$\rightarrow\mathtt{mg = 50}$

Hence rearranging 1 we get,

$\rightarrow\mathtt{N =mg(1 +\dfrac{ a}{g})}$

Now substituting the value of mg in  the above equation we get ,

$\rightarrow\mathtt{\orange{N =50(1 +\dfrac{ a}{g})}}$

The reading of the spring balance when lift is moving with uniform acceleration  in upward direction is 50(1+a/g) N