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2 years ago

10

A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.5 m. The block is released from

rest and undergoes simple harmonic motion. Calculate spring constant, mechanical force applied and velocity when x=15 cm.

1 answer:

morpeh [17]2 years ago

6 0

Answer:

0.7000cm

Explanation:

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In the periodic table, the properties repeat in what direction?

pogonyaev

Answer:

Left to right and top to bottom

Explanation:

On the periodic table, the properties repeat from left to right and from top to bottom.

Periodic properties have a pattern from the top to the bottom or down a group or family.

Also, across the period from left to right, they also show a repeating pattern.

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3 years ago

A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

$F = 14 lbf$

$F = 6.35 \times 9.8 N$

$F = 62.3 N$

stretch in the spring is given as

$x = 4 in = 0.1016 m$

now we will have

$F = kx$

$62.3 = k(0.1016)$

$k = 613.125 N/m$

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

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

$W = \frac{1}{2}(613.125)(0.254)^2$

$W = 19.8 J$

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3 years ago