it is a mechanics question???

Question

a load of mass 50kg stands on the horizontal floor of a goods lift which is accelerating vertically downwards at 2 m/s2.calculate the magnitude of the constant force exerted on the load by the floor.then a man who is in the lift ,ehich is still accelerating downwards at 2m/s2 ,pushes the load horizontally and the load is on the point of slipping ,given that the coefficient of friction between the load and the floor is 0.3, calculate the magnitude of the force exerted on the load by the man.

Answer 1

make the equation there is concept of psuedo force

mg-ma=N(normal contact force)

= (9.8 - 2) = 7.8m/s^2*50=390N

now u know that this contact force will b the constant force exerted on the load by floor

now u know that friction = u(meu)N
hence 0.3*390=117N

Answer 2

1)
F = ma
With respect to the load the floor is exerting an acceleration of (9.8 - 2) = 7.8m/s^2 upward.
The force acting on the load upward is

m x 7.8

50 x 7.8 = 390 N.

2)
The normal reaction is 390 N

F = µ m a

F = 0.3 x390 = 117 N

Answer 3

(1) f=400N

(2)friction=120N
force excerted=120N
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Answer 4

yes, it is. this is simply a newton's second law question.
f = ma