Answer:
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Answer:
net force acting on the floor is 100 kN
Explanation:
Given data:


dimension of floor = 2 m \times 0.5 m
we know that
Net force can be calculated as follow




Therefore net force acting on the floor is 100 kN
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
Answer:
The question is a problem that requires the principles of fracture mechanics.
and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.
Explanation:
Max Stress, σ = 2σ₀√(α/ρ)
where,
σ₀ = Tensile stress = 190MPa = 1.9x10⁸Pa
α = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m
ρ = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m
Max Stress, σ = 2 x 1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵
Max Stress, σ = 2 x 1.9x10⁸ x 6.708 Pa
Max Stress, σ = 2549MPa
Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa