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1 April, 07:32

7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surface and collides with a light spring-loaded guardrail with a spring constant k of 1.0 x 106 N/m. Find the maximum distance the spring is compressed.

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  1. 1 April, 07:48
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    0.505 m

    Explanation:

    From the question,

    The kinetic energy of the car = energy stored in the spring

    1/2mv² = 1/2ke² ... Equation 1

    Where m = mass of the car, v = velocity of the car, k = spring constant of the car, e = extension/compression

    make e the subject of the equation

    e = v√ (m/k) ... Equation 2

    We can calculate the value of v, by applying,

    v² = u²+2gH ... Equation 3

    Where u = initial velocity of the car, H = height of the car, g = acceleration due to gravity.

    Given: u = 0 m/s (from rest), H, 10 m, g = 9.8 m/s²

    Substitute into equation 2

    v² = 2 (10*9.8)

    v² = 196

    v = √196

    v = 14 m/s

    Also given: m = 1300 kg, e = 1.0*10⁶ N/m = 1000000 N/m

    Substitute into equation 2

    e = 14√ (1300/1000000)

    e = 14√ (0.00013)

    e = 14 (0.036)

    e = 0.505 m

    Hence the maximum distance of the spring is compressed = 0.505 m
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