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3 years ago

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The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.

1 answer:

Alika [10]3 years ago

8 0

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

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Answer:

a

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Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

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Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + \frac{1}{2}(-g)t^2$

=> $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

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Solving this using quadratic formula we obtain that

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Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

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and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

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Answer:

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