Troubleshooting Hook's Law for Balloon Placement in a Stretched Hose

[nemzy]

the question:

A balloon is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the strecthing of the hose obeys hooke's law with a spring constant 102 N/m. if the hose is stretched by 5.30m and then released, how much work does the force from the hose do on the baloon in the puch by the time the hose reaches its relaxed lenghty

The answer:

this is how i tried to solve it...F=-kx according to hooke's law..so i did -(102)(5.30)= -540.6 N ..and work is W=Fd..so i mulitiped the force with distance 5.30 and got -2865.18 which is not the answer...what am i doing wrong? thanks

 

[deltabourne]

Originally posted by nemzy
the question:

A balloon is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the strecthing of the hose obeys hooke's law with a spring constant 102 N/m. if the hose is stretched by 5.30m and then released, how much work does the force from the hose do on the baloon in the puch by the time the hose reaches its relaxed lenghty

The answer:

this is how i tried to solve it...F=-kx according to hooke's law..so i did -(102)(5.30)= -540.6 N ..and work is W=Fd..so i mulitiped the force with distance 5.30 and got -2865.18 which is not the answer...what am i doing wrong? thanks

See, force is varying in a spring.. meaning you can't use $W = F \times D$. Since work is the integral of force, the equation you want to use is:
$$W = \int_{0}^{x} F(x) dx$$
or
$$W = \frac{1}{2} k x^2$$

(In this case we integrate from x to 0, so we should remember to add a negative sign to our answer to show negative work)

 

[nemzy]

so its (1/2)kx^2 - (1/2)kx^2)

but since initial is 0 we get -(1/2)kx^2

and k=102 N/m and distance = 5.30 m

so -(1/2)(102)(5.30^2) = -1432.59 J

however -1432.59 J is not the right answer..am i missing something here?

 

[deltabourne]

Originally posted by nemzy
so its (1/2)kx^2 - (1/2)kx^2)

but since initial is 0 we get -(1/2)kx^2

and k=102 N/m and distance = 5.30 m

so -(1/2)(102)(5.30^2) = -1432.59 J

however -1432.59 J is not the right answer..am i missing something here?

what is the right answer?

 

[nemzy]

i have no idea..its those websites where u input the answer and if u get it wrong it tells u its wrong and when u get it right it tells u its right

 

[deltabourne]

Originally posted by nemzy
i have no idea..its those websites where u input the answer and if u get it wrong it tells u its wrong and when u get it right it tells u its right

I see.. have you tried rounding off to a different certain number of places? I know that did it for me last year for physics problems online when I used too many sigfigs

 

[Doc Al]

Originally posted by nemzy
however -1432.59 J is not the right answer..am i missing something here?

The work done by the hose is positive since the force and displacement are in the same direction.

 

[nemzy]

thx!