Physics help

[Gibson486]

1. An elevator cable breaks when a 735 kg elevator is 22.5 m above the top of a huge spring (k = 8.00 x10^4 N/m) at the bottom of the shaft. Calculate the amount the spring compresses. (Note that here work is done by both the spring and gravity.)

I keep getting .09m. If work is done by the spring, how do you calculate its work?

2.A 5.0 kg block is pushed 8.0 m up a rough 38° inclined plane by a horizontal force of 65 N. If the initial speed of the block is 1.7 m/s up the plane and a constant kinetic friction force of 28 opposes the motion, calculate the the final kinetic energy of the block.

I got 3.38, but my friends tell me it is wrong.

3. What should be the spring constant k of a spring designed to bring a 1420 kg car to rest from a speed of 111 km/h so that the occupants undergo a maximum acceleration of 5.0 g (1 g = 9.8)?

4. A 1200 kg car rolling on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 3 m. What is the spring constant of the spring?

I keep getting 7173 and 1420

 

[Koing]

Originally posted by: Gibson486
1. An elevator cable breaks when a 735 kg elevator is 22.5 m above the top of a huge spring (k = 8.00 x10^4 N/m) at the bottom of the shaft. Calculate the amount the spring compresses. (Note that here work is done by both the spring and gravity.)

I keep getting .09m. If work is done by the spring, how do you calculate its work?

2.A 5.0 kg block is pushed 8.0 m up a rough 38° inclined plane by a horizontal force of 65 N. If the initial speed of the block is 1.7 m/s up the plane and a constant kinetic friction force of 28 opposes the motion, calculate the the final kinetic energy of the block.

I got 3.38, but my friends tell me it is wrong.

3. What should be the spring constant k of a spring designed to bring a 1420 kg car to rest from a speed of 111 km/h so that the occupants undergo a maximum acceleration of 5.0 g (1 g = 9.8)?

4. A 1200 kg car rolling on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 3 m. What is the spring constant of the spring?

I keep getting 7173 and 1420

Hey can you post and label all the formulas you have. I did Physics last year and remember how to most of these questions but can't remember the exact formulas off hand so put them down and I could work some out pretty quick for you.

 

[Gibson486]

W=fd cos theta
w= change in Kenetic energy
KE= .5mv^2
f= kx

 

[Koing]

Originally posted by: Gibson486
W=fd cos theta
w= change in Kenetic energy
KE= .5mv^2
f= kx

I meant the spring constant one. its has length of the spring and a square root. I can't remember how this one exactly goes.

 

[Gibson486]

F=kx, k being spring constant, while F is force and x is compression.

 

[Yossarian]

I will do 1 & 3 because I don't feel like drawing a FBD.

1. F = kx
mg = kx
x = mg/k
x = (735)(9.8)/80000 = 0.09 m

W = Fx
W = (735)(9.8)(0.09) = 648.5 J


3. 1st find the acceleration distance.

v1^2-v0^2 = 2ax

Since v1=0, x = -(v0^2) / (2a)

Don't forget to convert v0 to m/s (30.83 m/s)

x = -(30.83^2) / [-(2*5*9.8)]
x = 9.7 m

The car's entire kinetic energy will be changed to potential energy in the spring.

KE = PE
0.5mv0^2 = 0.5kx^2

Solve for k, plug in numbers, k = 14,345 N/m

 

[Gibson486]

.09m is wrong for #1

 

[Yossarian]

oh sorry I didn't see the part about the elevator being above the spring. Just find the velocity of the elevator after a 22.5m fall, figure the KE, then convert that to PE for the spring.

 

[Gibson486]

Is there anyway to do the problems w/o potential energy?

 

[eLiu]

uh...work done by a spring...integrate f=-kx, and get:
work=1/2kx0^2 - 1/2kxf^2

x0=x initial
xf=x final

tho, i usually just drop the sign and figure out what sign the answer "should" be at the end...yeah i'm a bad student.

To do #1, you do: 1/2mv^2 = 1/2kx^2 - mgx
v=speed when it hits the spring
x=how far the spring compresses

 

[raptor13]

.09m is wrong for the first problem for the following reason:


Gravity acts on the elevator at all times including while it is compressing the spring. You have to set the work of the spring equal to the work done by gravity on the elevator as it moves from 22.5m + deltaX of the spring.



Now do it yourself. I already took this class.

 

[raptor13]

For the second problem, calculate the component of the 65N force acting in the direction of the plane. Subtract 28N (friction always opposes motion, so you subtract). Then subtract the component of the boxes weight acting along the plane. You now have a net force, an initial speed, and a distance over which it acts.

Work = Force * Distance = Change in Kinetic Energy


You can calculate intial kE, you just got delta(kE), you can get final kE. And you know kE = .5mv^2. Solve for v, you're done.




EDIT: I have my own homework to do now so you're on your own for the rest. They both operate on exactly the same principles of these first two problems, though, so you should be able to get them. If you really want, you can PM and I'll get back to you later but I hope you can figure them out. Good luck!

 

[Rahminator]

Number one is easy. All initial energy is potential, and all final energy of the system is elastic. Because the energy of this system can't change (no external forces), therefore mgh = .5kx^2. I get 2.01 m.

 

[Rahminator]

I'm bored so I will do some more of these. I don't undersand the wording of 2 (horizontally as in horizontally to paper or parallel to incline plane surface?), so I will start with 3.

3. All kinetic energy of the car is converted to elastic potential energy of the spring, so:
(1/2)mv^2 = Fx = ma(x)

mass cancels out and they give you maximum decelartion the car can sustain, so solving for x (remember to convert velocity to m/s by dividing by 3.6):
x = v^2 / (2a)

plug in numbers and you get spring compression of 9.7 m. Since all KE of car gets converted in PE(spring), you can use this equation:
(1/2)mv^2 = (1/2)kx^2

(1/2) factor cancels out and by solving for k you get:
k = mv^2 / x^2.
Plug in your numbers and you get 14344.7 N / m.

4. Again, KEi gets converted PEsf:
(1/2)mv^2 = 1/2kx^2
(1/2) cancels out. Solve for k and you get:
k = mv^2 / x^2
Plug in numbers (convert velocity to m/s) and you get 37037 N/m.

 

[Gibson486]

1) this was a tricky one, but I got 2.10m. Can you guys check it out? Remember, the spring also does work, so what i did was included the work of the work of th spring and used f= kx and got .09m, then I added it to the 2.01 I got. total 2.10.

2) I am still stuck on finding the kenetic energy. I did the sum of W= K(f) - K(i), but I get a negitive K(f)! From my understanding, Kenetic energy cannot be negitive. I need help on this.

3) I got 3586.2 N/m.

Here is what I did

1/2kx^2=1/2mv^2, the 1/2 cancels out.
kx^2= mv^2.
I need K and I need x, so I sepereted the left side into kx * x.
I can now use F= KX, however, again, I need KX, but this time, I need to find force.
The max acceration is 5g's (5*9.8) so that equals 49
Therefore, kx = 49
then, 49 x = mv^2
I solve for X.
Then, the X value I get, I plug back into the equation and solve for K.

 

[Gibson486]

BTW, 4 is not 14344.7

Thanks. I really appreciate your guys help.