f=mg equation (1 Viewer)

[davidbarnes]

"In a particular 'bungee' jump, a 75kg person 'freefalls' for 5 seconds (ignore air resistance) before the bungee brings the person to a stop in a further 3 seconds. Determinev the average deceleration force acting during this time."

You can only use the equation F=mg to sole the above problem. The answer is "-1225N", although could it also be 49 ms^-1 which is the answer I obtained. If not, why not?

 

[veloc1ty]

"In a particular 'bungee' jump, a 75kg person 'freefalls' for 5 seconds (ignore air resistance) before the bungee brings the person to a stop in a further 3 seconds. Determinev the average deceleration force acting during this time."

You can only use the equation F=mg to sole the above problem. The answer is "-1225N", although could it also be 49 ms^-1 which is the answer I obtained. If not, why not?

Uh, force is always measured in Newtons isn't it? And it's asking for the average deceleration force. :\

 

[Forbidden.]

Uh, force is always measured in Newtons isn't it? And it's asking for the average deceleration force. :\

F is in Newtons (N or kg.ms^-2)
m is in kilograms (kg)
g is in metres per second squared (ms^-2)

Proposed equations were:

F = mg - ma
F = mg + ma
F = m(v - u)/t where a_av = v - u / t

BTW, is that the only values given in the question, 75kg, 3s and 5s ?

 

[davidbarnes]

Yes those are the only values. That is the exact question above.

 

[Leo 100]

force= mass x gravity
are you sure this is the only equation you can use?
why would they give you time?

gravity = 9.8

 

[xiao1985]

v = u + at

u = 0, a = -9.8, t = 5
v = -49 ms^-1

Now: the person is stopped in 3 seconds. so average deceleration is:

a = -49/3 ms^-2

and note: F = m a

so average deceleration force is -49/3 x 75 = -1225N

 

[davidbarnes]

force= mass x gravity
are you sure this is the only equation you can use?
why would they give you time?

gravity = 9.8

Yes, sheet specifically tells you to use that and only that equation.

 

[Forbidden.]

v = u + at

u = 0, a = -9.8, t = 5
v = -49 ms^-1

Now: the person is stopped in 3 seconds. so average deceleration is:

a = -49/3 ms^-2

and note: F = m a

so average deceleration force is -49/3 x 75 = -1225N

Why didn't I think of this >=\