Answer:
For block A, a = 9.66 ft/s²
For block B, a = 15 ft/s²
Explanation:
A free body diagram for this force system is attached to this solution
Mass of block A = m₁ = 8 lb
Mass of block B = m₂ = 6 lb
Coefficient of kinetic friction = μ
Normal reaction on the blocks = N
Spring stiffness of the spring btw block A and B = k = 20 lb/ft
Compression of the spring = 0.2 ft
Analysing Block A first
The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 8 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 8/32.2 = 0.248 lbm
(20×0.2) - (0.2 × 8) = (0.248) aₓ
aₓ = 9.66 ft/s²
The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 6 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 6/32.2 = 0.186 lbm
(20×0.2) - (0.2 × 6) = (0.186) aₓ
aₓ = 15 ft/s²
To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.
Density can be defined as
Where
m = Mass
V = Volume
For state one we know that
For state two we have to
Therefore the total change of mass would be
Therefore the mass of air that has entered to the tank is 6.02Kg