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kirill [66]
3 years ago
11

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

pring and
A) Calculate how much it would stretch if the same person was lying on it.
B) How much it would stretch if the person jumped from 35 m
Physics
1 answer:
posledela3 years ago
4 0

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

The net would strech 0.03167 m

If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

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7 0
3 years ago
a 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m
antiseptic1488 [7]

Answer:

(a) 0 J

(b) 828.96 J

(c) 828.96 J

(d) 828.96 J

(e) 0 J

(f) 828.96 J

Explanation:

Given:

Mass of the rock is, m=3.18\ kg

Initial height of the rock is, h_{i}=26.6\ m

Final height of the rock is, h_f=0\ m

Initial velocity of the rock is, v_i=0\ m/s(Rest)

Acceleration due to gravity is, g=9.8\ m/s^2

(a)

Initial kinetic energy is given as:

K_i=\frac{1}{2}mv_i^2

Plug in the given values and solve for 'K_i'. This gives,

K_i=\frac{1}{2}\times 3.18\times 0^2=0

Therefore, initial kinetic energy is 0 J.

(b)

Initial potential energy is given as:

U_i=mgh_i\\U_i=3.18\times 9.8\times 26.6=828.96\ J

Therefore, initial potential energy is 828.96 J.

(c)

Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,

ME=K_i+P_i\\ME=0+828.96=828.96\ J

Therefore, mechanical energy at 26.6 m is 828.96 J.

(d)

Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.

Therefore, kinetic energy at the 0 m is given as:

K_f=U_i=828.96\ J

(e)

Potential energy at the final height is given as:

P_f=mgh_f=3.18\times 9.8\times 0=0\ J

Therefore, final potential energy is 0 J.

(f)

Total mechanical energy is a constant and doesn't depend on the height.

So, total mechanical energy at 0 m is same as that at 26.6 m.

Total mechanical energy is 828.96 J.

8 0
3 years ago
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