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2 years ago

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A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the

car has a speed of 25.6 m/s. Calculate the mechanical energy of the car at the top and bottom of the hill. (Assume the bottom of the hill has a height of 0 m, g=9.80 ms2/).

1 answer:

rjkz [21]2 years ago

3 0

Explanation: Solution

1.

Gravitational potential energy

U=mgh=500*9.8*50

U=245000 J

2.

Kinetic energy is present at bottom of the hill

K=(1/2)mV2=(1/2)*500*27.82

K=193210 J

3.

Work done by friction

W=193210-245000=-51790 J

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As we know that train is initially moving with the speed

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

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I Hope this helps, greetings ... Dexteright02! =)

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