A spring gun with a 75 N/m spring constant is loaded with a 5 g foam dart and isaimed vertically. When the spring is compressed by 10 cm and

Question

A spring gun with a 75 N/m spring constant is loaded with a 5 g foam dart and isaimed vertically. When the spring is compressed by 10 cm and then released, the fireddart rises to a max height of 5 m above the end of the spring gun. Assuming the dartexperiences a constant friction force due to the air, how fast is it traveling when ithas fallen 2 m from its maximum height

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Verity 6 months 2021-07-17T21:08:50+00:00 1 Answers 7 views 0

Answers ( )

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    2021-07-17T21:10:04+00:00

    Answer:

    The speed is  v = 4.425 m/s

    Explanation:

    From the question we are told that

         The spring constant is  k = 75 \ N /m

          The mass of the foam dart is m =  5 g =  \frac{5}{100} = 0.05 \ kg

          The compression distance is  d = 10 cm =  0.1 m

           The height which the gun raised the dart is  h = 5 m

            The change in height is  \Delta h =  2 m

            The new height is h_2 = 5 -2 = 3 m

    Generally from the law of conservation of energy

                E_s = KE

    Where E_s is the energy stored in spring and it is mathematically represented as

                E_s =  \frac{1}{2} k d^2

      KE is the kinetic energy possessed by the dart when it is being shut and this is mathematically represented as

                  KE = \frac{1}{2} mv^2_r

    So

              \frac{1}{2} k d^2 =  \frac{1}{2} mv^2_r

    Substituting values

              0.5 * 75 * 0.1 =  0.5 * 0.0005 * v^2_r

    =>     v_r   =   \sqrt{\frac{0.5 * 75 * 0.1}{0.5 * 0.0005 } }

            v_r  =  12.25 m/s

    When the dart is at  the maximum height the

         let it acceleration due air resistance be z

    So by equation of motion

              v^2 =  u^2 - 2ah

    Where v is the velocity at maximum height which is equal to zero

        and  u is it initial velocity before reaching maximum height which we calculated as v_r  =  12.25 m/s

           and a is the acceleration due to gravity + the acceleration due to air resistance

         So

              a =  z+g

                 =  9.8 + z

    =>    v^2 =  u^2 - 2(9.8 +z)h

    Substituting values

              0 =  12.25^2 - 2(9.8 +z)h

    Making z the subject

              z =  \frac{ 12.25}{2 * 5}  - 9.8

             z =  \frac{ 12.25}{2 * 5}  - 9.8

             z = 5 m/s

    When the dart is moving downward we can mathematically represent the motion as

            v^2 = u^2 + 2ah

    Since the motion is downward and air resistance is upward we have that

             a =  g – z

    and the the initial velocity u becomes the velocity at maximum height

    i.e u = 0

         And v is the velocity the dart has when it is moving downward

                   So

                             v^2 = 0 + 2 * (g -z )h

    Substituting values

                            v = \sqrt{0+ 2 (10 - 5) * 2}

                            v = 4.425 m/s

                 

                   

       

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