Answer:
(a) Spring constant = 1850 N/m.
(b) Energy stored in the spring = 9.25 J.
(c) Muzzle velocity of the dart = 43.01 m/s.
Explanation:
(a) Spring constant
From hook's law,
F = ke ...................... Equation 1
Where F = force on the gun, k = spring constant of the gun, e = extension of the gun's spring.
Making k subject of the equation,
k = F/e ............... Equation 2
Given: F = 185 N, e = 10 cm = 0.1 m.
Substitute into equation 2
k = 185/0.1
k = 185/0.1
k = 1850 N/m.
Hence the spring constant = 1850 N/m.
(b) Energy stored in the spring,
E = 1/2ke²............... Equation 3
Where E = Energy stored in the spring, k = spring constant, e = extension.
Given: k = 1850 N/m, e = 0.1 m.
Substitute into equation 3
E = 1/2(1850)(0.1)²
E = 925(0.01)
E = 9.25 J.
(c) Muzzle velocity of the dart.
Kinetic energy of the dart = 1/2mv²
Note: The kinetic energy of the dart is equal to the energy stored in the sprig.
E = 1/2mv²
Where m = mass of the dart, v = velocity of the dart.
Making v the subject of the equation,
v = √(2E/m)............... Equation 4
Given: E = 9.25 J, m = 10 g = 0.01 kg
Substituting these values into equation 4
v = √(2×9.25/0.01)
v = √(18.5/0.01)
v = √1850
v = 43.01 m/s.
