How does newtons laws apply to the egg drop experiment
1 answer:
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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
The maximum length during the motion is
Explanation:
From the question we are told that
The mass is
The vertical spring length is
The unstretched length is
The initial speed is
The new length of the spring
The spring constant k is mathematically represented as
Where F is the force applied
y is the difference in weight which is
The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)
Now substituting values accordingly
The elastic potential energy is given as
where D is this the is the displacement
Since Energy is conserved the total elastic potential energy would be
Substituting value accordingly
So to obtain total length we would add the unstretched length
So we have