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A stuntman of mass 48 kg is to be launched horizontally out of a spring-
loaded cannon. The spring that will launch the stuntman has a spring
coefficient of 75 N/m and is compressed 4 m prior to launching the
stuntman. If friction and air resistance can be ignored, what will be the
approximate velocity of the stuntman once he has left the cannon?
O A. 5 m/s
O B. 21 m/s
O C. 9 m/s
D. 13 m/s

Sagot :

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The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

Velocity:

This is the ratio of displacement to time. The S.I unit of Velocity is m/s.  The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

  • mv²/2 = ke²/2
  • mv² = ke².................. Equation 1

⇒ Where:

  • m = mass of the stuntman
  • v = velocity of the stuntman
  • k = force constant of the spring
  • e = compression of the spring

⇒ Make v the subject of the equation

  • v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

  • m = 48 kg
  • k = 75 N/m
  • e = 4 m

⇒ Substitute these values into equation 2

  • v = √[(75×4²)/48]
  • v = √25
  • v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: https://brainly.com/question/10962624

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