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3 years ago

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If the work done to stretch an ideal spring by 4.0 cm is 6.0 J, what is the spring constant (force constant) of this spring? A)

300 N/m B) 3000 N/m C) 3500 N/m D) 7500 N/m E) 6000 N/m

1 answer:

Ostrovityanka [42]3 years ago

6 0

Answer:

D) 7500 N/m

Explanation:

Given :

Work done by the spring, W = 6 J

Stretch in the spring, x = 4 cm = 0.04 m

Consider k be the spring constant.

The relation between work done and stretch of the spring is:

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

Substitute the suitable values in the above equation.

$6=\frac{1}{2}k(0.04)^{2}$

$k=\frac{12}{1.6\times10^{-3} }$

k = 7500 N/m

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Answer:

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Explanation:

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Answer:

Part a)

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Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

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Part a)

differentiate x and y two times with respect to time to find the acceleration

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Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

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$\theta = -137.7 degree$

Part c)

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7 0

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