A .001 kg bead slides without friction around a loop-the-loop. The bead is released from a height of 20.6 m from the bottom of t
1 answer:
Answer:
11.37 m/s
Explanation:
We are given;
Mass; m = 0.001 kg
Initial velocity; v1 = 0 m/s (since it was released from rest)
Initial height; h1 = 20.6m
From the attached image of the bead in the loop, we can see that height at A; h2 = twice the radius = 2R = 2 × 7 = 14 m
g = 9.8
To solve for the speed at point, we will use formula for conservation of energy. Thus;
½m(v1)² + mgh1 = ½m(v2)² + mgh2
Divide through by m to get;
½(v1)² + gh1 = ½(v2)² + gh2
Plugging in the relevant values;
½(0²) + (9.8 × 20.6) = ½(v2)² + (9.8 × 14)
Rearranging, we have;
½(v2)² = ½(0²) + (9.8 × 20.6) - (9.8 × 14)
Simplifying, we have;
½(v2)² = 64.68
Multiply both sides by 2 to get;
(v2)² = 2 × 64.68
(v2)² = 129.36
v2 = √129.36
v2 = 11.37 m/s
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Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s