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Zepler [3.9K]
3 years ago
15

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring wi

th a spring constant of 35.0 N/m.35.0 N/m. When he pulls the slingshot back 0.375 m0.375 m from a nonstretched position, it just does not launch its payload as far as he wants. His physics professor "helps" by telling him to aim for an elastic potential energy of 22.0 J.22.0 J. Tom decides he just needs elastic bands with a higher spring constant. By what factor does Tom need to increase the spring constant to hit his potential energy goal?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

8.9

Explanation:

We can start by calculating the initial elastic potential energy of the spring. This is given by:

U=\frac{1}{2}kx^2 (1)

where

k = 35.0 N/m is the initial spring constant

x = 0.375 m is the compression of the spring

Solving the equation,

U=\frac{1}{2}(35.0)(0.375)^2=2.5 J

Later, the professor told the student that he needs an elastic potential energy of

U' = 22.0 J

to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:

k'=\frac{2U'}{x^2}=\frac{2(22.0)}{0.375^2}=313 N/m

Therefore, Tom needs to increase the spring constant by a factor:

\frac{k'}{k}=\frac{313}{35}=8.9

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A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

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The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!

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