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2 years ago

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The potential energy of an object attached to a spring is 2.60 J at a location where the kinetic energy is 1.40 J. If the amplit

ude ???? of the simple harmonic motion is 16.0 cm, calculate the spring constant ???? and the magnitude of the largest force ????spring, max that the object experiences.
????= Nm

????spring, max= N

1 answer:

anyanavicka [17]2 years ago

6 0

Answer:

This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.

U = (1/2)kx^2

U = (1/2)(5.3)(3.62-2.60)^2

U = 2.75706 J

Read more on Brainstorm - httpd://brainstorm/question/7981441#read more

Explanation:

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<h3>Relative error</h3>

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

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Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

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3 years ago