college physics?

Question

A 65-kg. trampoline artist jumps vertically upward with a speed of 5.0 m/s. If the trampoline behaves like a spring with spring stiffness constant 62000 N/m, how far does he depress it?

Answer 1

7.6 feet

Answer 2

At the point of maximum depression (d) the potential energy (PE) of the depressed tramp will equal the kinetic energy (KE) of the artist just as she leaves the tramp having a spring coefficient = k = 62 X 10^3 N/m.

Thus, PE = kd^2 = 1/2 mv^2 = KE; so that d^2 = (1/2 mv^2)/k = mv^2/(2 k) and d = sqrt(mv^2/(2k))...m = 65 kg, k = 62 X 10^3 N/m, and v = 5 m/sec..you can do the math.

Important physics lesson, in general, the conservation of energy law will use the fact that KE converts to PE, and vice versa. The problems get more complex when that conversion is not 100%; that can occur when there are other forces at work like friction forces. In this case, some of that potential energy could be converted into heat as well as kinetic energy; so KE < PE would be true or KE + Q = PE; where Q is that heat energy..

Answer 3

I think you can do this one simpl;y by considering energy.

the KE of the jumper when she hits the tramp is 1/2 mv2 = 1/2 65 * 5^2
(she has to have the same velocity downwards at the end of the jump as she had upwards to begin with)

the PE of a compressed spring is 1/2 kx2
and should ultimately equal the KE:

1/2 65 25 = 1/2 62000 x2
.0262 = x2
x=.161 meter, about 6 inches.

Answer 4

I think that you can solve this one by balancing the Energy. Since he was 65kg and 5m/sec then using mv^2/2 will give you his kinetic energy and then using the spring potential energy = kx^2/2 where k = spring constant and x = distance depressed. You can solve for x.

I get about x = .16meters

Answer 5

who the hell gave me a thumbs down for this answer?