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A toy dart gun uses a spring with a spring constant of 35 N/m. To use the dart gun, you compress the spring by pushing in a dart of mass 0.002 kg.
a. If you compress the spring by 0.02 m, what force is the spring exerting on the dart?

b. With the spring compressed 0.02 m, how much elastic potential energy is stored in the spring?

c. If you release the spring, it pushes the dart forward. What is the kinetic energy of the dart when it reaches the natural length of the spring? Explain how you found the answer.

d. What is the speed of the dart when it reaches the natural length of the spring?

Respuesta :

onyebuchinnaji

(a) The force is the spring exerting on the dart is 0.7 N.

(b) The elastic potential energy stored in the spring is 7 x 10⁻³J.

(c) The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

(d) The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

Force the spring exerting on the dart

F = kx

where;

  • k is spring constant
  • x is compression of the spring

F = 35 N/m x 0.02 m

F = 0.7 N

Energy stored in the spring

U = ¹/₂kx²

U = ¹/₂(35)(0.02)²

U = 7 x 10⁻³ J

Kinetic energy of the dart

The elastic potential energy of the spring will be converted into kinetic energy of the dart, with a magnitude of  7 x 10⁻³ J when it reaches natural length of the spring.

Speed of the dart

K.E = ¹/₂mv²

v² = (2K.E)/m

where;

  • v is the speed of the dart
  • m is mass of the dart

v² = (2 x 7 x 10⁻³)/(0.002)

v² = 7

v = 2.65 m/s

Thus, the force is the spring exerting on the dart is 0.7 N.

The elastic potential energy stored in the spring is 7 x 10⁻³J.

The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

Learn more about elastic potential energy here: https://brainly.com/question/25996974

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