Atoolbox of mass 3.2kg is lowered by a rope from the roof to the ground. find the acceleration of the toolbox when the force of the rope is: a) 40.0 n b) 20.0n
a) 2.7m/s^2, up b) 3.6m/s^2, down
i would like an explanation or equation to get to that answer. you!

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

                              a + g = F / m

                              a = F/m - g

a)                            a = 40/3.2 - 9.81

                              a = 2.69 ≈ 2.7 m/s²   positive up

b)                            a = 20/ 3.2 - 9.81

                              a = 6.25 - 9.81

                                  = - 3.56 ≈ - 3.6 m/s²  negative down

vapour or gas assuming atmospheriv pressure