A 78 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.4m/s.
v=7.7m/s
If the trampoline behaves like a spring of spring constant 5.6×104 N/m , how far does he depress it?
When does v = 7.7 m/s?
To find the distance the trampoline depresses, we need to use the principle of conservation of mechanical energy.
The initial mechanical energy of the system is equal to the final mechanical energy.
The initial mechanical energy is the kinetic energy of the trampoline artist just before he jumps, given by:
KE_initial = (1/2) * mass * velocity^2
Substituting the given values, we have:
KE_initial = (1/2) * 78 kg * (4.4 m/s)^2
= 457.536 J
The final mechanical energy is the potential energy of the trampoline artist when he reaches the highest point of his jump, given by:
PE_final = (1/2) * k * displacement^2
where k is the spring constant and displacement is the distance the trampoline depresses.
Substituting the given value of the spring constant, we have:
PE_final = (1/2) * (5.6×10^4 N/m) * displacement^2
Since the mechanical energies are equal, we have:
KE_initial = PE_final
457.536 J = (1/2) * (5.6×10^4 N/m) * displacement^2
Simplifying the equation, we get:
displacement^2 = (2 * KE_initial) / k
displacement^2 = (2 * 457.536 J) / (5.6×10^4 N/m)
displacement^2 = 0.01629 m^2
Taking the square root of both sides, we find:
displacement = √(0.01629 m^2)
displacement ≈ 0.1277 m
Therefore, the trampoline depresses approximately 0.1277 meters.