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2 February, 19:54

A toy gun uses a spring to project a 5.9-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gun is 17 cm long, and a constant frictional force of 0.035 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.7 cm for this launch?

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  1. 2 February, 21:48
    0
    Stored energy in spring = 1/2 k x², k is spring constant, x is compression.

    = 1/2 x 8 x (5.7 x 10⁻²) ²

    = 129.96 x 10⁻⁴ J

    Energy lost due to friction = force x distance

    =.035 x. 17

    =.00595 J

    Energy used in providing kinetic energy to projectile.

    129.96 x 10⁻⁴ -.00595

    .012996 -.00595

    =.007046 J

    So

    1/2 m v² =.007046

    v² =.007046 x 2 /.0059

    = 2.3885

    v = 1.545 m / s
  2. 2 February, 23:01
    0
    mass, m = 5.9 g = 0.0059 kg

    Spring constant, K = 8 N/m

    length of barrel, s = 17 cm = 0.17 m

    frictional force, f = 0.035 N

    Compression in spring, Δs = 5.7 cm = 0.057 m

    let the speed of the projectile is v.

    Energy stored in the spring, E = 0.5 KΔs²

    E = 0.5 x 8 x 0.057 x 0.057 = 0.012996 J

    Work done by the friction force, W = f x s = 0.035 x 0.17 = 0.00595 J

    Energy used to move the projectile, E' = E - W

    E' = 0.012996 - 0.00595 = 0.007046 J

    E' = 1/2 mv²

    0.007046 = 0.5 x 0.0059 x v²

    v² = 2.3885

    v = 1.55 m/s
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