The 102N acting on the ropes being pulled by eric and kim have some of that force acting horizontally, and some of it vertically. By visualizing it as a right angled triangle, with the hypotenuse the length of the diagonal force, and each side the length of the horizontal and vertical forces, you can use trigonometry to calculate the length of the vertical force. You are told that it is at an angle of 30 with the vertical rope, therefore you know the length of the hypotenuse, and the angle between it and the vertical force, so using trig: (vertical force=x)
x/102=cos(30)
x=102*cos(30)
x=88.33
Therefore the diagonal ropes give a vertical force of 88.33N, and the centre rope, as it acts vertically, gives a vertical force of all 102N. The total:
88.33*2+102=278.66N
I don't know if this is very clear, I hope its good enough to help. If you don't understand, just ask, and I can answer any questions!!! :)
The block's speed at the point where x=0.25A is v = 31.95 cm/s.
<h3>What is Spring constant?</h3>
The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.
question is incomplete, this is the remaining statement
What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?
x = Asin(wt)
v = Aw coswt
at t = 0
w = sqrt(k/m)
v = Aw
A = v/w
A = 7.17 cm
part b )
E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
mv^2 + k(1/4A)^2 = 1/2kA^2
mv^2 + kA^2/16 = kA^2
mv^2 = kA^2 - kA^2/16
mv^2 = 15kA^2/16
v^2 = 15/16 * (k/m) * A^2
v^2 = 15/16 *w^2A^2
v = sqrt(15/16) * wA
v = 31.95 cm/s
to learn more about spring constant go to -
brainly.com/question/23885190
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