A body of mass 1 kg is suspended from a spring balance calibrated for acceleration due to gravity of 10 mcdot s^(-2). What is the reading on the spring balance when the system (i) is ascending with an acceleration of 5 mcdot s^(-2) and (ii) is descending with the same acceleration? [ g = 10 m cdot s^(-2)]

A body of mass 1 kg is suspended from a spring balance calibrated for

1.	A body of mass 1 kg is suspended from a spring balance calibrated for acceleration due to gravity of 10 mcdot s^(-2). What is the reading on the spring balance when the system (i) is ascending with an acceleration of 5 mcdot s^(-2) and (ii) is descending with the same acceleration? [ g = 10 m cdot s^(-2)]
Answer» Solution :Reaction force on the balance when it ascends, R = m(g + a ) = 1(10 + 5) = 15 N. `therefore` In this CASE, the reading of the spring balance ` = (15)/(10) = 1.5` kg. (ii)Reaction force when the system descends, R = m(g - a) = 1(10 - 5) = 5 N. `therefore` In this case, the reading of the spring balance `= (5)/(10)` = 0.5 kg,