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jehan4141
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Please help me with this problem. many thanks :)
A 55-kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion. Find the force (magnitude and direction) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.6 m/s^2. Ignore the effects of air resistance.
This is my work and answer. Other students are getting all sorts of different answers than me. Am I doing something wrong? Or is my work correct? Thank you so much!
The forces acting upon her are gravity downward, and the force of the bungee cord (Fb) which pulls her back in the upward direction.
Fnet = Fb - mg = ma
We know that she is experiencing an acceleration of 7.6 m/2^s in the downward direction. Thus:
Fb - mg = m(-a)
Fb = m(-a) + mg
Fb = (55)(-7.6) + 55(9.8)
Fb = 121 N upward
Thank you again for looking!
A 55-kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion. Find the force (magnitude and direction) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.6 m/s^2. Ignore the effects of air resistance.
This is my work and answer. Other students are getting all sorts of different answers than me. Am I doing something wrong? Or is my work correct? Thank you so much!
The forces acting upon her are gravity downward, and the force of the bungee cord (Fb) which pulls her back in the upward direction.
Fnet = Fb - mg = ma
We know that she is experiencing an acceleration of 7.6 m/2^s in the downward direction. Thus:
Fb - mg = m(-a)
Fb = m(-a) + mg
Fb = (55)(-7.6) + 55(9.8)
Fb = 121 N upward
Thank you again for looking!