Question

An object of mass 3.0 kg is attached to the hook of a spiral balance and the balance is suspended from the roof of a lift. What is the reading on the spring balance when the lift goes down with an acceleration of 0.1 m/s2​

Answer 1

The force acting on the object is the weight of the object, which is given by:

F = m * g

where m = 3.0 kg is the mass of the object and g = 9.81 m/s^2 is the acceleration due to gravity.

F = 3.0 kg * 9.81 m/s^2 = 29.43 N

The net force acting on the object when the lift goes down is the difference between the weight of the object and the force required to accelerate it downward:

Fnet = F - m * a

where a = 0.1 m/s^2 is the acceleration of the lift.

Fnet = 29.43 N - 3.0 kg * 0.1 m/s^2 = 29.13 N

The reading on the spring balance is equal to the net force acting on the object, which is 29.13 N.