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olga55 [171]
2 years ago
10

A 70.0 kg man jumping from a window lands in an elevated fire rescue net 11.0 m below the window. He momentarily stops when he h

as stretched the net by 1.50 m. Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by 1.50 m. (10 pts)
Physics
1 answer:
levacccp [35]2 years ago
6 0

Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

Let K is the spring constant.

In the equilibrium position

mg = K l

70 x 9.8 = K x 11

K = 62.36 N/m

Potential energy stored, U = 0.5 x K x Δl²

U = 0.5 x 62.36 x 1.5 x 1.5

U = 70.15 Joule

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Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

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Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

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Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

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In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

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