A 60.0-kg person drops from rest a distance of 1.20 m to a platform of negligible mass supported by an ideal stiff spring of neg
ligible mass. the platform drops 6.00 cm before the person comes to rest. what is the spring constant of the spring?
1 answer:
Draw a diagram to illustrate the problem s shown in the figure below.
At state A, the person has potential energy relative to the spring of
PE = (60 kg)*(9.8 m/s²)*(1.2 m)
= 705.6 J
At state B, the PE is converted to kinetic energy which is equal to the potential energy.
Assume that aerodynamic losses are ignored during the free fall.
At state C, the potential energy (or kinetic energy) is converted into strain energy in the spring. Assume that energy losses are negligible.
The spring deflects by 6 cm = 6x10⁻² m, and its spring constant = k N/m.
Therefore
(1/2)(k N/m)(6x10⁻² m)² = 705.6 J
k = 705.6/0.03 = 23520 N/m = 23.52 kN/m
Answer: 23.52 kN/m
At state A, the person has potential energy relative to the spring of
PE = (60 kg)*(9.8 m/s²)*(1.2 m)
= 705.6 J
At state B, the PE is converted to kinetic energy which is equal to the potential energy.
Assume that aerodynamic losses are ignored during the free fall.
At state C, the potential energy (or kinetic energy) is converted into strain energy in the spring. Assume that energy losses are negligible.
The spring deflects by 6 cm = 6x10⁻² m, and its spring constant = k N/m.
Therefore
(1/2)(k N/m)(6x10⁻² m)² = 705.6 J
k = 705.6/0.03 = 23520 N/m = 23.52 kN/m
Answer: 23.52 kN/m
You might be interested in
<h2>Answer: 117.626m/s</h2>
Explanation:
The escape velocity is given by the following equation:
(1)
Where:
is the Gravitational Constant and its value is
is the mass of the asteroid
is the radius of the asteroid
On the other hand, we know the density of the asteroid is and its volume is
.
The density of a body is given by:
(2)
Finding :
(3)
(4) This is the mass of the spherical asteroid
In addition, we know the volume of a sphere is given by the following formula:
(5)
Finding :
(6)
(7)
(8) This is the radius of the asteroid
Now we have all the necessary elements to calculate the escape velocity from (1):
(9)
Finally:
This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.
Answer:
719
Explanation:
Conversion
1 picometer (pm) is equivalent to meter
1 micrometer is equivalent to meter
To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by meter we get that the number of sheets will be as follows
Therefore, they are approximately 719 sheets
<h2>Answer:</h2>
<em>Hello, </em>
<h3><u>QUESTION)</u></h3>Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².
<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>
- Δt = (√2xgxh)/9,8
- Δt = (14√10)/9,8
- Δt ≈ 4,5 s