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dem82 [27]
3 years ago
14

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) surfa

ce 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.
Physics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

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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Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, L=50\ mH=0.05\ H

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So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

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