A 45 kg gymnast steps off a platform located 1.3 m above a spring board. The spring board compresses 25 cm. a)What is the spring constant of the spring board? b) If energy is in fact conserved in this problem, explain how high the gymnast will spring upward. Assume the gymnast does not use is legs to propel himself upward. c) Suppose a diving board has a spring constant of 9500 N/m. If a 75 kg diver compresses the diving board a distance of 35 cm, how high will the diver be launched above the relaxed position of the diving board?
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a) To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The formula for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the spring (in Newtons)
k is the spring constant (in N/m)
x is the displacement from the equilibrium position (in meters)
In this case, the gymnast's weight will be balanced by the force exerted by the spring, so we can set up the equation:
mg = kx
Where:
m is the mass of the gymnast (in kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the given values:
m = 45 kg
g = 9.8 m/s^2
x = 25 cm = 0.25 m
We can solve for k:
k = (mg) / x
k = (45 kg * 9.8 m/s^2) / 0.25 m
k ≈ 1764 N/m
Therefore, the spring constant of the spring board is approximately 1764 N/m.
b) If energy is conserved in this problem, the potential energy gained by compressing the spring will be converted entirely into kinetic energy when the gymnast springs upward.
The potential energy stored in a spring is given by:
PE = (1/2) kx^2
Where:
PE is the potential energy (in Joules)
In this case, the potential energy gained by compressing the spring is equal to the potential energy the gymnast will have while springing upward. So:
PE = (1/2) kx^2
Plugging in the given values:
k ≈ 1764 N/m
x = 0.25 m
PE = (1/2) * 1764 N/m * (0.25 m)^2
PE ≈ 55.125 Joules
The potential energy gained will be converted entirely into kinetic energy, so:
KE = PE = 55.125 Joules
The kinetic energy of an object is given by:
KE = (1/2) mv^2
Where:
m is the mass of the object (in kg)
v is the velocity of the object (in m/s)
Since we are assuming the gymnast does not use their legs to propel themselves upward, the velocity will be determined by the conversion of potential energy to kinetic energy.
Solving for v:
55.125 Joules = (1/2) * 45 kg * v^2
v^2 = (2 * 55.125 Joules) / 45 kg
v^2 ≈ 2.45 m^2/s^2
v ≈ 1.56 m/s
Therefore, the gymnast will spring upward with a velocity of approximately 1.56 m/s.
To find how high the gymnast will spring upward, we can use the equation for gravitational potential energy:
PE = mgh
Where:
PE is the potential energy (in Joules)
m is the mass of the object (in kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height above the reference position (in meters)
Since the kinetic energy gained from the compressed spring will be entirely converted into potential energy, we can set up the equation:
KE = (1/2) mv^2 = mgh
Solving for h:
(1/2) mv^2 = mgh
h = (1/2) v^2 / g
Plugging in the given values:
v ≈ 1.56 m/s
g = 9.8 m/s^2
h = (1/2) * (1.56 m/s)^2 / 9.8 m/s^2
h ≈ 0.1256 m
Therefore, the gymnast will spring upward to a height of approximately 0.1256 m.
c) Using the same equation for potential energy, we can calculate the height the diver will be launched above the relaxed position of the diving board.
Plugging in the given values:
m = 75 kg
g = 9.8 m/s^2
k = 9500 N/m
x = 35 cm = 0.35 m
h = (1/2) * k * x^2 / (m * g)
h = (1/2) * 9500 N/m * (0.35 m)^2 / (75 kg * 9.8 m/s^2)
h ≈ 0.0625 m
Therefore, the diver will be launched above the relaxed position of the diving board to a height of approximately 0.0625 m.
a) To find the spring constant of the spring board, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, the gymnast's weight (mg) is equal to the force exerted by the spring (F). The displacement (x) can be calculated as the initial height (1.3 m) minus the amount the spring board compresses (25 cm, converted to meters by dividing by 100).
Therefore, we have:
mg = -kx
Plugging in the values:
(45 kg)(9.8 m/s^2) = -k(1.3 m - 0.25 m)
Now we can solve for the spring constant, k:
k = -((45 kg)(9.8 m/s^2)) / (1.3 m - 0.25 m)
b) If energy is conserved in this problem, we can use the principle of conservation of mechanical energy to find how high the gymnast will spring upward.
The mechanical energy before the spring is compressed consists of the potential energy due to the gymnast's height above the relaxed position of the spring board. The mechanical energy after the spring is compressed consists of the potential energy due to the gymnast's new height above the relaxed position of the spring board, as well as the elastic potential energy stored in the spring.
Since the gymnast does not use their legs to propel themselves upward, we can assume that all the initial potential energy is transferred to the final potential energy and the elastic potential energy.
The initial potential energy can be calculated as the mass of the gymnast times the acceleration due to gravity times the height above the relaxed position of the spring board, which is 1.3 m.
The final potential energy can be calculated as the mass of the gymnast times the acceleration due to gravity times the new height above the relaxed position of the spring board.
The elastic potential energy can be calculated as (1/2) times the spring constant times the square of the displacement of the spring.
By setting the initial potential energy equal to the final potential energy plus the elastic potential energy, we can solve for the final height.
c) Using the same principles as part b, we can calculate the final height of the diver above the relaxed position of the diving board.
The initial potential energy can be calculated as the mass of the diver times the acceleration due to gravity times the height above the relaxed position of the diving board, which is 0 (since the diver starts from a relaxed position).
The final potential energy can be calculated as the mass of the diver times the acceleration due to gravity times the new height above the relaxed position of the diving board.
The elastic potential energy can be calculated as (1/2) times the spring constant times the square of the displacement of the spring.
By setting the initial potential energy equal to the final potential energy plus the elastic potential energy, we can solve for the final height.