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3 years ago

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A spring 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.

1 answer:

kobusy [5.1K]3 years ago

8 0

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.

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