Answer:
The effective spring constant of the firing mechanism is approximately 21.804 N/m
Step-by-step explanation:
The mass of the circus performer, m = 75.6 kg
The angle of elevation above the horizontal, θ = 37.2°
The instrument the cannon uses to propel the performer = Strong elastic band
The length by which the band stretches, x = 2.70 m
The height of the performer above the floor at the point the performer flies free of the bands = The height of the net into which he is shot
The time he takes to travel the horizontal distance between the point he is shot and the net, t = 2.34 s
The horizontal distance between the the point he is shot and the net, dₓ = 26.9 m
Therefore, the horizontal component of the velocity, vₓ, can be found with the following kinematic equation of motion;
vₓ = dₓ/t
∴ vₓ = 2.70 m/(2.34 s) = (15/13) m/s
By the resolution of the initial velocity, 'v', into the horizontal and vertical component, we have;
The horizontal component, vₓ = v × cos(θ)
∴ v = vₓ/(cos(θ))
v = (15/13)/(cos(37.2°)) ≈ 1.45
The initial velocity, v ≈ 1.45 m/s
The kinetic energy given to the circus performer by the elastic band, K.E., is given as follows;
K.E. = (1/2)·m·v²
Where;
m = The mass of the performer = 75.6 kg
v = The velocity given to the performer ≈ 1.45 m/s
∴ K.E. ≈ (1/2) × 75.6 × 1.45² ≈ 79.475
∴ K.E. ≈ 79.475 J
The kinetic energy, K.E., given to the performer, (1/2)×m×v² = The kinetic energy of the elastic band = (1/2)×k×x²
Where;
k = The effective spring constant of the firing mechanism
x = The extension of the sling = 2.70 m
Therefore, we get;
K.E. = (1/2)·m·v² = (1/2)·k·x²
∴ K.E. ≈ 79.475 J = (1/2) × 75.6 × 1.45² = (1/2) × k × (2.70 m)²
∴ 79.475 J ≈ (1/2) × k × (2.70 m)²
k = 79.475 J/((1/2) × (2.70 m)²) ≈ 21.804 N/m
The effective spring constant of the firing mechanism, k ≈ 21.804 N/m.