Aspring operated dart gun fires 10 g darts. arming the gun requires 185 n of force and results in the shortening of the spring by 10 cm. find the spring constant. find the energy stored in the spring. find the muzzle velocity of the dart.

(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.