A 65kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0m/s. a) How fast is he going as he lands on the trampoline, 3.0m below. b) if the trampoline behaves like a spring with a spring stiffness constant 6.2*10^4N/m, how far does he depress it?
.307
To answer part a) of the question, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the trampoline artist is equal to the sum of kinetic energy and potential energy.
The kinetic energy (KE) of the artist is given by the equation KE = (1/2) * m * v^2, where m is the mass of the artist (65kg) and v is the velocity (5.0m/s).
KE = (1/2) * 65kg * (5.0m/s)^2
= 812.5J
The potential energy (PE) of the artist when he lands on the trampoline is given by the equation PE = m * g * h, where g is the acceleration due to gravity (9.8m/s^2) and h is the height (3.0m).
PE = 65kg * 9.8m/s^2 * 3.0m
= 1914J
Since the mechanical energy is conserved, the initial mechanical energy (812.5J) is equal to the final mechanical energy (KE + PE) when the artist lands on the trampoline.
812.5J = KE + PE
812.5J = KE + 1914J
812.5J - 1914J = KE
KE = -1101.5J (negative because the artist is moving downward)
Now, we can use the kinetic energy equation to find the final velocity (vf) of the artist when he lands on the trampoline:
KE = (1/2) * m * vf^2
-1101.5J = (1/2) * 65kg * vf^2
Solving for vf:
vf^2 = (-1101.5J * 2) / 65kg
vf^2 = -67.8m^2/s^2 (negative because the artist is moving downward)
vf = ± √(-67.8m^2/s^2)
vf ≈ ± 8.24m/s
Therefore, the magnitude of his velocity as he lands on the trampoline is approximately 8.24m/s.
To answer part b) of the question, we can use Hooke's Law to determine the amount the trampoline is depressed. Hooke's Law states that the force required to compress or extend a spring is directly proportional to the displacement from its equilibrium position.
The formula for the potential energy stored in a spring is given by the equation PE = (1/2) * k * x^2, where k is the spring stiffness constant (6.2 * 10^4N/m) and x is the displacement of the spring from its equilibrium position.
PE = (1/2) * (6.2 * 10^4N/m) * x^2
Since the potential energy stored in the spring is equal to the difference in potential energy before and after the artist jumps on the trampoline, we can set up the equation:
PE = m * g * h = (1/2) * (6.2 * 10^4N/m) * x^2
Simplifying, we can solve for x:
(1/2) * (6.2 * 10^4N/m) * x^2 = 65kg * 9.8m/s^2 * 3.0m
(6.2 * 10^4N/m) * x^2 = 19140N
x^2 = 19140N / (6.2 * 10^4N/m)
x^2 ≈ 0.308m^2
Taking the square root, we find:
x ≈ ±√(0.308m^2)
x ≈ ±0.555m
Therefore, the trampoline depresses by approximately 0.555m when the artist jumps on it.
To solve for the final velocity you need the time and the point where his velocity was zero.
v=v0+at since your solving for a vf=0
v0/a=t
5/9.8=0.51s
Then solve for his highest position using the time interval
x=x0+v0t+1/2(at^2)
x=3+(5)(0.51)+1/2(-9.8)(0.51^2)
=4.28m
Next, solve use energy equations setting the potential energy at the top of his jump equal to the kinetic energy when he hits the trampoline.
1/2(mv^2)=mgh
1/2(v^2)=gh
v=sqrt(2gh)
v=sqrt(2(9.8)(4.28))=9.2m/s
A 57kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7m/s. How fast is he going as he lands on the trampoline 3.8m below?
a) When hitting the trampoline H=3.0 m below, the kinetic energy will have increased
from (1/2)MVo^2 to
(1/2)MV1^2 =
(1/2)MVo^2 + MgH
Solve for V1
V1^2 = Vo^2 + 2gH
b) (1/2)MV1^2 = (1/2)kX^2
Solve for X, the trampoline displacement.