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Zepler [3.9K]
4 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]4 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 small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
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Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

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\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

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