Respuesta :
Answer:
a) 100 kJ
b) 90.909 × 10³ N
c) 190.08 kJ
Explanation:
Given:
Initial velocity, v₁ = 50 m/s
Final velocity, v₂ = 0
Mass of the pilot, m = 80 kg
Distance covered in snow, d = 1.1 m
Now,
Work done (W) = change in kinetic energy
or
W = [tex]\frac{1}{2}m(v_1^2-v_2^2)[/tex]
in substituting the values, we get
W = [tex]\frac{1}{2}\times80 (50^2-0^2)[/tex]
or
W = 100000 J = 100 kJ
b) Average force exerted by snow
Now, Force = (change in momentum/time taken)
or
F = ΔM/Δt
now,
ΔM = m(v₁ - v₂)
Δt = distance/ (average speed) = 1.1/[(50 + 0)/2] = 0.044 s
F = 80 × (50-0)/0.044
F = 90.909 × 10³ N
c) Work done by air resistance
W = change in energy by the air resistance
now, the change is energy is as:
from potential energy at the height, H = 370 m to the potential energy acquired just before touching the snow
thus,
W = mgH - (1/2)mv₁²
on substituting the values, we get
W = 80 × 9.8 × 370 - (1/2) × 80 × 50² = 190080 J = 190.08 kJ
A) The work done by the snow in bringing him to rest is; 100000 J
B) The average force exerted on him by the snow to stop him is; F = 90909.09 N
C) Work done on him by the air resistance is;
W_s = 190362 J
We are given;
Mass of pilot; m = 80 kg
Terminal velocity; v = 50 m/s
A) Let us first estimate the change in kinetic energy which is workdone by the snow in bringing him to rest;
W = ∆K = Final kinetic energy - initial kinetic energy
∆K = ½mv² - ½mu²
u is initial velocity = 0 m/s
Thus;
∆K = ½(80 × 50²)
∆K = 100000 J
B) The average force exerted on him by the snow to bring him to stop is;
F = W/∆x
We are given ∆x = 1.1 m
Thus;
F = 100000/1.1
F = 90909.09 N
C) Final velocity without drag is;
v = √(2gx)
We are given x = 370 m. Thus;
v = √(2 × 9.8 × 370)
v_f = 85.2 m/s
Terminal velocity is velocity without drag. Thus, the work done on him by the air resistance as he fell is;
W_s = ½m((v_f)² - (v²))
W_s = ½ × 80 × (85.2² - 50²)
W_s = 190362 J
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