[zenite]

A 65.5 kg firefighter slides down a pole while a constant frictional force of 300 N retards her motion. A horizontal 20.0 kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest 3.30 m above the platform, and the spring constant is 4000 N/m.

(a) Find the firefighter's speed just before she collides with the platform.
(b) Find the maximum distance the spring is compressed. (Assume the frictional force acts during the entire motion.)

part a is 5.88m/s

part b, i am stuck. i got an answer, but its wrong. I used the conservation of momentum to calculate the velocity of both the platform and the firefighter, form an energy equation, using KE of Platform and Firefighter, Ugravitational and Uspring. somehow, I got it wrong. can anyone help?

[freedoms]

May i know how to solve for the first part?

[PECTOPAH]

Originally Posted by zenite

A 65.5 kg firefighter slides down a pole while a constant frictional force of 300 N retards her motion. A horizontal 20.0 kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest 3.30 m above the platform, and the spring constant is 4000 N/m.

(a) Find the firefighter's speed just before she collides with the platform.
(b) Find the maximum distance the spring is compressed. (Assume the frictional force acts during the entire motion.)

part a is 5.88m/s

part b, i am stuck. i got an answer, but its wrong. I used the conservation of momentum to calculate the velocity of both the platform and the firefighter, form an energy equation, using KE of Platform and Firefighter, Ugravitational and Uspring. somehow, I got it wrong. can anyone help?

show ur workings and we can see wad went wrong

[Jodar]

Originally Posted by zenite

A 65.5 kg firefighter slides down a pole while a constant frictional force of 300 N retards her motion. A horizontal 20.0 kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest 3.30 m above the platform, and the spring constant is 4000 N/m.

(a) Find the firefighter's speed just before she collides with the platform.
(b) Find the maximum distance the spring is compressed. (Assume the frictional force acts during the entire motion.)

part a is 5.88m/s

part b, i am stuck. i got an answer, but its wrong. I used the conservation of momentum to calculate the velocity of both the platform and the firefighter, form an energy equation, using KE of Platform and Firefighter, Ugravitational and Uspring. somehow, I got it wrong. can anyone help?

[1] - When the compression of the spring is maximised, ALL the energy from the motion of the object is converted into elastic potential energy of the spring.

[2] - We know that Hooke's Law states that "Within the limit of proportionality, the extension of a spring is directly proportionate to the force acting on it", i.e. if you plot F against x (extension), you will get Work Done = (1/2)(F)(x), since work done is area under F-x graph.

[3] - Since Hooke's Law says F = kx, Work Done = (1/2)(k)(x)², where k is given: 4000N/m.

If you can get part a, it means you can get the amount of energy the body has when it just touches the platform. That energy, minus the gravitational potential energy lost (through the compression) and the energy loss due to friction, will give you the elastic potential energy of the spring at its maximum compression.

Equate all that and solve for the unknown x, which is the maximum distance the spring is compressed.

[freedoms]

May i know how to calculate the first part?

My wquation for the 1st part was:

Potential Engery = Kinetic Energy

(65.5 X 10 X 3.3 ) - (300 X 3.3 ) = 1/2 X 65.5 X V^2

Erm this is O level physics right?

[Jodar]

Originally Posted by freedoms

May i know how to calculate the first part?

My wquation for the 1st part was:

Potential Engery = Kinetic Energy

(65.5 X 10 X 3.3 ) - (300 X 3.3 ) = 1/2 X 65.5 X V^2

Erm this is O level physics right?

Sorry, buddy. It's A-level physics. Well, you can solve the first part using O-level concepts, but you do require A-level knowledge to complete the second part. Anyway, your method is right, 'cept we assume gravity to be 9.81 instead of 10 in the A's.

[typeX]

Originally Posted by Jodar

[1]

[2] - We know that Hooke's Law states that "Within the limit of proportionality, the extension of a spring is directly proportionate to the force acting on it", i.e. if you plot F against x (extension), you will get Force = (1/2)(k)(x).

um... just wondering, isn't Hooke's law F = k*x? where did the 1/2 come from? =/

[kaitke]

mind telling whats the answer for part b?
nt sure if i'm right, but

firstly i equate f=ma, ---> (65.5)(9.81)-300=(65.5)a in order to find a which is the acceleration.
where i will then get a = 5.23. Then, since F=Ke, and i have K which is 4000, but need to find e, i must first have the value of F. So to find F, we equate F= (65.5)(5.23)+20(9.81). Afterwhich, we sub it back into the F=Ke equation to find e which is then 0.135.

[Jodar]

Originally Posted by typeX

um... just wondering, isn't Hooke's law F = k*x? where did the 1/2 come from? =/

Thanks for pointing it out. It should be Work Done = (1/2)(F)(x), where F = kx. Sorry, too many things on my mind.

[zenite]

ok, thx for all the reply. I will post my working here, pls let me know what went wrong


By conservation of energy,
Summation of Work = Change in KE
Change in U(gravitational) + Change in U(internal) = K2 - K1
Ug = U1 - U2 = (M+m)gh - 0 = (M+m)gh = (65.5+20)*9.81*x = 839x ... (h=x)
U(int) = 1/2(kx^2) = 0.5*4000*x^2 = 2000x^2
K2 - K1 = K2 - 0 = 1/2((M+m)vf^2) = 1/2*(65.5+20)*v^2

now, to find v. v is the velocity of the platform and the firefighter. using the model for inelastic collision, momentum is conserved.

pi = pf
mvi = (m+M)vf
65.5*5.88 = (65.5+20)*vf
vf = 4.5m/s

back to energy...
839x + 2000x^2 = 1/2*(85.5)*4.5^2
839x + 2000x^2 - 866 = 0

Using general eqn...
x = 0.481mm

Can anyone verify if this is correct. I am unable to since the quiz is closed.

zenite added 6 Minutes and 21 Seconds later...

Originally Posted by kaitke

mind telling whats the answer for part b?
nt sure if i'm right, but

firstly i equate f=ma, ---> (65.5)(9.81)-300=(65.5)a in order to find a which is the acceleration.
where i will then get a = 5.23. Then, since F=Ke, and i have K which is 4000, but need to find e, i must first have the value of F. So to find F, we equate F= (65.5)(5.23)+20(9.81). Afterwhich, we sub it back into the F=Ke equation to find e which is then 0.135.

I think its wrong. you are equating the F from the falling firefighter to the F from the spring. Force is not conserved. And the platform acceleration is not 9.81, it must be equal to the acceleration of the firefighter. (how can they have different a when they are together).

Also, a is ONLY constant when there is no change in F. For a spring, F increases as x (compression) increase. So F = ma is not possible to solve part b since F is changing. Using the Work-Kinetic theory will be a better approach.

[kaitke]

erm i think yr mistaken, how can the acceleration be the same for the platform and the fireman? my working of F= (65.5)(5.23)+20(9.81), the 20(9.81) is in fact actually the weight of the platform, and the acceleration of the firemen will definitely be 5.23 if u use 65.5)(9.81)-300=(65.5)a, where i have already taken into account the resistance he faced by deducting 300. And anyway, it should be assumed that the total energy gets converted over to elastic potential energy, as said by Jodar as well earlier on ''[1] - When the compression of the spring is maximised, ALL the energy from the motion of the object is converted into elastic potential energy of the spring.''

And another point to note, Ug = U1 - U2 = (M+m)gh - 0 = (M+m)gh = (65.5+20)*9.81*x = 839x ... (h=x). ===> this portion is wrong already, u can't put it this way because u did not take into account of the resistive force.