Asked by yo
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)
Answers
Answered by
drwls
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
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
Answered by
yo
Forgot to mention is that the answer for unit is J
Answered by
drwls
Spring constants cannot have units of J. It has to have force/length units, not force x length.
Neither can Velocities.
Neither can Velocities.
Answered by
adasd
Weight = m*g
700 = m*g
h = 12m
Ep = m*g*h
Ep = 700 * 12
Ep = 8400 J
700 = m*g
h = 12m
Ep = m*g*h
Ep = 700 * 12
Ep = 8400 J
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