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Elenna [48]
2 years ago
10

A spring with a mass attached to it is stretched from the rest position of 20 cm to the position of 87 cm. If the spring constan

t of this spring is 1729 N/m, what is the potential energy stored in this spring?
561.29 J
120.97 J
388.07 J
254.79 J
Physics
2 answers:
Marianna [84]2 years ago
8 0

Answer:

388.07 J

Explanation:

Rina8888 [55]2 years ago
6 0

Answer:

C) 388.07

Explanation:

It is a spring therefore, there must be Elastic potential energy. Now what is the formula for elastic potential energy? It is 1/2 *k *x^2

K=spring constant

x= displacement  

1/2=.5 (hate fractions)

Now, Displacement simply means (how much was it moved?) So, that means subtract.

87-20= 67   x=67

Next, do x^2 (so you don't make a mistake by squaring everything :)  )

67*67=4,489

NOW, do....

.5*1729*4,489=3,880,740.5

Now convert to Joules, I just did 3,880,740.5  / 10^3 = 3880.745

round to 3880.7 J  

                                                                ~There you go loves !

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<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

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x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

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<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

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a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

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