Answer:
Step-by-step explanation:
Part A:
Using conservation of energy, we can find the speed of the trampoline artist just before he lands on the trampoline. At the top of the platform, he has gravitational potential energy equal to mgh, where m is his mass, g is the acceleration due to gravity, and h is the height of the platform. At the bottom of his jump, he has kinetic energy equal to (1/2)mv^2, where v is his speed just before he lands on the trampoline. We can equate these two energies and solve for v:
mgh = (1/2)mv^2
v = sqrt(2gh)
where h = 2.0 m and g = 9.81 m/s^2. Plugging in the values, we get:
v = sqrt(2(9.81 m/s^2)(2.0 m)) = 6.26 m/s
Therefore, the trampoline artist is going 6.26 m/s as he lands on the trampoline.
Part B:
The force exerted by the trampoline on the artist is equal to the weight of the artist, which is mg, where g is the acceleration due to gravity. This force causes the trampoline to compress a distance d, which we want to find. From Hooke's law, we know that the force exerted by a spring is equal to its spring constant times its deformation from its equilibrium length. Therefore:
mg = kd
where k is the spring constant of the trampoline. Solving for d, we get:
d = (mg)/k
where m = 75 kg, g = 9.81 m/s^2, and k = 6.7×10^4 N/m. Plugging in the values, we get:
d = (75 kg)(9.81 m/s^2)/(6.7×10^4 N/m) = 0.109 m
Therefore, the trampoline depresses 0.109 m when the artist lands on it.