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3 years ago

What is the spring potential energy of a spring that is stretched 15 cm if its spring constant is 350 n/m?

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

Spring potential energy = 7.875Nm

Explanation:

<u>Given the following data;</u>

Extension, e = 15cm to meters = 15/100 = 0.15m

Spring constant, k = 350n/m

To find the force;

Force = spring constant * extension

Force = 350 * 0.15

Force = 52.5 Newton.

Now to find the spring potential energy we would use the formula below;

Spring potential energy = force * extension

Substituting into the equation, we have;

Spring potential energy = 52.5 * 0.15

<em>Spring potential energy = 7.875Nm</em>

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4) 9.6 m/s^2

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$F_f = \mu mg=(0.16)(31.0 kg)(9.8 m/s^2)=48.6 N$

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5) 462.3 N

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Answer:

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Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

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So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

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Work done = change in kinetic energy

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3 years ago