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.

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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-scientists plan in going deeper than 120 meters

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hope this !