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Vanyuwa [196]
3 years ago
10

A 1200 kg car rolling on a horizontal surface has speed v = 50 km/h when it strikes a horizontal coiled spring and is brought to

rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
Mathematics
1 answer:
tangare [24]3 years ago
8 0
When the car is moving, it has Kinetic Energy. As soon as the car entercount with the spring, its Kinetic Energy will transfer to Elastic Potential Energy. 

<span>KE = EPE </span>
<span>1/2mv^2 = 1/2kx^2 </span>

<span>change the speed into m/s </span>
<span>50 km/hr = 13.9 m/s </span>

<span>1/2(1200)(13.9)^2 = 1/2k(2.2)^2 </span>

<span>now you do the math to solve for k.
k= 115926/2.42
K= 47903.30</span>
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\frac{9}{10} \div \frac{6}{7}
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6 0
3 years ago
The bases of the trapezoid are 6 and 10. The height is 4. Find the area.
kari74 [83]
<h2>Answer:</h2>

b. \ \boxed{32 \sq. \ units}

<h2>Step-by-step explanation:</h2>

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6 0
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4 0
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irinina [24]
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3 years ago
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