Physics, 09.08.2019 22:30 kpnclark3920

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. the car goes up a 12--high hill, then descends 20 to the track's lowest point. you've determined that the spring can be compressed a maximum of 2.3 and that a loaded car will have a maximum mass of 410 . for safety reasons, the spring constant should be 15 larger than the minimum needed for the car to just make it over the top.
(a) what spring constant should you specify?
(b)what is the maximum speed of a 370 car if the spring is compressed the full amount?

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Answer from: kprincess16r

(a) Based on the safety reasons, the spring constant to be specified is 1761.97 N/m.

(b) The speed of 370 kg car when the spring is compressed fully is 5.02 m/s.

The given parameters;

distance up the hill, s₁ = 12 mdistance down the hill, s₂ = 20 mextension of the spring, x = 2.3 mmass of the car, m = 410 kg

The weight of the car is calculated as follows;

W = mg

W = 410 x 9.8 = 4018 N

The spring constant is calculated as;

F = kx

$k = \frac{F}{x} = \frac{4018}{2.3} \\\\k = 1746.97 \ N/m$

Based on the safety reasons, the spring constant to be specified is calculated as;

k = 1746.97 + 15 = 1761.97 N/m

(b)

The speed of 370 kg car when the spring is compressed fully is calculated as;

$\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v= \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{1761.97\times 2.3^2}{370} } \\\\v =5.02 \ m/s$

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Answer from: eaalvarezelisa01

Part a)

$k_{eq} = 2.74 \times 10^5 N/m$

Part b)

v = 63.8 m/s

Explanation:

Part a)

A car can go maximum 12 m high hill and then come back to other end at the depth of 20 m

So here by energy conservation we can say

Energy stored in the spring = gravitational potential energy at the top

$\frac{1}{2}kx^2 = mgH$

$\frac{1}{2}k(2.3)^2 = (410 kg)(9.81 m/s^2)(12 m)$

$k = 1.82 \times 10^4 N/m$

now spring constant is 15 times larger than this value

so we have

$k_{eq} = 2.74 \times 10^5 N/m$

Part b)

When mass of the car is m = 370 kg

then the highest speed of the car when it will reach the lowest point is given as

$\frac{1}{2}kx^2 + 0 = \frac{1}{2}mv^2 - mgh$

$\frac{1}{2}(2.74 \times 10^5)(2.3^2) = \frac{1}{2}(370)v^2 - (370 kg)(9.81)(8m)$

$7.54 \times 10^5 = 185 v^2$

v = 63.8 m/s

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

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