A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for
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Answer:
19.8 J
Explanation:
According to the law of conservation of energy, the total mechanical energy of the spring (sum of kinetic energy and elastic potential energy) must be conserved:
(1)
where we have
is the initial kinetic energy of the spring, which is zero because the spring starts from rest (2)
is the elastic potential energy of the spring when it is fully stretched
is the kinetic energy of the spring when it reaches the natural length
is the elastic potential energy of the spring when it reaches its natural length, which is zero because the stretch in this case is zero (3)
So
where
k = 440 N/m is the spring constant
is the initial stretching of the spring
Substituting,
And so using eq.(1) and keeping in mind (2) and (3) we find
In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
