Respuesta :
Ball hits the floor at 1.6 m below the initial position
So here we can say that
[tex]\Delta y = -1.6 m[/tex]
also it will hit at horizontal distance of 5.2 m
[tex]\Delta x = 5.2 m[/tex]
let say its velocity in x and y directions are given as
[tex]v_x ,v_y[/tex]
now we can say
[tex]v_x* t = 5.2[/tex]
[tex]-1.6 = v_y * t - \frac{1}{2}gt^2[/tex]
from above two equations
[tex]-1.6 = v_y* \frac{5.2}{v_x} - 4.9t^2[/tex]
as we know that it is projected at an angle of 35 degree
so we know that
[tex]v_y = v_x tan35[/tex]
[tex]-1.6 = 5.2 tan35 - 4.9 t^2[/tex]
[tex]4.9 t^2 = 3.64 + 1.6 = 5.24[/tex]
[tex]t = 1.03 s[/tex]
now we have
[tex]v_x * 1.03 = 5.2 [/tex]
[tex]v_x = 5.03 m/s[/tex]
also we can find vertical component of velocity as
[tex]v_y = v_x tan35 = 3.52 m/s[/tex]
now we will find net velocity as
[tex]v^2 = v_x^2 + v_y^2[/tex]
[tex]v^2 = 5.03^2 + 3.52^2[/tex]
[tex]v = 6.14 m/s[/tex]
now by energy conservation we can say
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]
[tex]k*(0.18)^2= 0.032*(6.14)^2[/tex]
[tex]k = 37.2 N/m[/tex]
so spring constant is 37.2 N/m
Answer:
The value of spring constant is [tex]37.23\;\rm{N[/tex].
Explanation:
Given: A compressed spring launches a [tex]32\;\rm{g[/tex] metal ball at a [tex]35[/tex] degree angle compressing the spring [tex]18\;\rm{cm[/tex] cause the ball to hit the floor [tex]1.6\;\rm{m[/tex] below the point at which it leaves the spring after traveling [tex]5.2\;\rm{m[/tex] horizontally.
As mentioned in question:
The ball hits the floor at [tex]1.6\;\rm{m[/tex] below the initial position then [tex]\Delta}y=-1.6\;\rm{m[/tex]
And horizontal distance of ball is [tex]\Delta{x=5.2\;\rm{m[/tex].
Now, velocity in both directions are [tex]v_x\;\&\;v_y[/tex].
[tex]v_x\times t=5.2\\-1.6=v_y\times t-\frac{1}{2}gt^2\\-1.6=v_xtan35^\circ\times \frac{5.2}{v_x}-4.9t^2\\4.9t^2=3.64+1.6\\[/tex]
[tex]t^2=\frac{5.24}{4.9}\\[/tex]
[tex]t=\sqrt{1.06938}\\t=1.03\;\rm{sec[/tex]
Then, [tex]v_x=\frac{5.2}{1.03}=5.03\;\rm{m/s[/tex] and [tex]v_y=5.03tan35^\circ=3.52\;\rm{m/s[/tex]
Net velocity[tex](v)=v_x^2+v_y^2[/tex]
[tex]v^2=(5.03)^2+(3.52)^2\\v^2=37.6913\\v=6.14\;\rm{m/s[/tex]
Using Energy conservation: [tex]\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]
[tex]k(0.18)^2=0.032(6.14)^2\\[/tex]
[tex]k=\frac{1.2063}{0.0324}\\k=37.23\;\rm{N[/tex]
Hence, The value of spring constant is [tex]37.23\;\rm{N[/tex].
Learn more about Energy conservation law here:
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