A 700-N man jumps out of a window into a fire net 10 m below. The net stretches 2 m before bringing the man to rest and tossing him back into the air. What is the spring constant of the net? What is the velocity just before hitting the net? (hint: Use cons. of energy twice)

Answer 1

The man's mass is
M = 700/g = 71.4 kg

When fully stretched 2 meters, gravitational potential energy of
M*g*(10 + 2) = 117.6 J is converted to potential energy of the net, which is
(1/2) k X^2, where X = 2 meters.
k = 2*117.6/X^2 = 59 N/m

Before hitting the net, hewill have fallen H = 10 meters, so that
V = sqrt(2*g*H) = 14 m/s

Answer 2

Forgot to mention is that the answer for unit is J

Answer 3

Spring constants cannot have units of J. It has to have force/length units, not force x length.

Neither can Velocities.

Answer 4

Weight = m*g
700 = m*g

h = 12m

Ep = m*g*h
Ep = 700 * 12
Ep = 8400 J