Answer:
A
[tex]x = 0.198456 \ m [/tex]
B
[tex]h = 1.3061 \ m [/tex]
C
[tex] v = 5.06 \ m/s [/tex]
D
[tex]d = 4.0273 \ m [/tex]
Explanation:
Considering the first question
From the question we are told that
The spring constant is [tex]k = 32.50 N/m[/tex]
The potential energy is [tex]PE = 0.640 \ J[/tex]
Generally the potential energy stored in spring is mathematically represented as [tex]PE = \frac{1}{2} * k * x^2[/tex]
=> [tex]0.640= \frac{1}{2} * 32.50 * x^2[/tex]
=> [tex]x = \sqrt{0.03938}[/tex]
=> [tex]x = 0.198456 \ m [/tex]
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation
[tex]PE = mgh[/tex]
=> [tex]0.640 = 0.050 * 9.8 * h[/tex]
=> [tex]h = 1.3061 \ m [/tex]
Considering the third question
The height at which the dart was fired horizontally is [tex]H = 3.90\ m[/tex]
Generally from the law of energy conservation
[tex]PE = KE [/tex]
Here KE is kinetic energy of the dart which is mathematical represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
=> [tex]0.640 = \frac{1}{2} * 0.050 * v^2 [/tex]
=> [tex] v^2 = 25.6 [/tex]
=> [tex] v = 5.06 \ m/s [/tex]
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as
[tex]t = \frac{ 2 * H }{g}[/tex]
=> [tex]t = \frac{ 2 * 3.90 }{9.8 }[/tex]
=> [tex]t = 0.7959 \ s [/tex]
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as
[tex]d = v * t[/tex]
=> [tex]d = 5.06 * 0.7959[/tex]
=> [tex]d = 4.0273 \ m [/tex]
