1) Data:
Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m
2) Questions: V at Y = 3m and X at Y = 3 m
3) Calculate components of the initial velocity
Vox = Vo * cos(37°) = 15.97 m/s
Voy = Vo * sin(37°) = 12.04 m/s
4) Formulas
Vx = constant = 15.97 m/s
X = Vx * t
Vy = Voy - g*t
Y = Yo + Voy * t - g (t^2) / 2
5) Calculate t when Y = 3m (first time)
Use g ≈ 9.8 m/s^2
3 = 12.04 * t - 4.9 t^2
=> 4.9 t^2 - 12.04t + 3 = 0
Use the quadratic equation to solve the equation
=> t = 0.28 s and t = 2.18s
First time => t = 0.28 s.
6) Calculate Vy when t = 0.28 s
Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s
7) Calculate V:
V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s
tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°
Answer: V ≈ 18.5 m/s, with angle ≈ 30°
8) Calculate X at t = 0.28s
X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m
Answer: X ≈ 4,5 m
Answer:
Whether the force exerted by the locomotive on the wall was larger
Than the force the locomotive could exert on the wall.
Explanation:
The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.
So we have;
F12=-F21
F12 is the force in a direction
-F21 is the force in the opposite direction.
Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.
If the springs are connected together from end to end, they are arranged in series. For springs in series, the forces are additive.
Spring 1: F1 = k1(Δx1)
Spring 2: F2 = k2(Δx2)
Spring 1: F3 = k3(Δx3)
Total Force = k1(Δx1)+k2(Δx2)+k3(Δx3)
Total Force = (k1+k2+k3)(Δx,total)
The spring constants are added together and multiplied with the total length of elongation to find the total force acting on it.
Answer:
The string is more likely to break when the circle is vertical
Explanation:
The string is more likely to break when the circle is vertical because tension in the string is the same throughout the circular motion and it is equal to mrw^2 when it is rotated in the horizontal plane while tension in the string is maximum and is given by mrw^2+mg when in the vertical plane.
