Need help with Newton's Law of Physics

[alex740rider]

I've been stuck on part b for several hours now (feels like the whole day).

1. Homework Statement :

2. Homework Equations :
2 k * x sin(θ') = 100/g * a
sum(Fy) = F net
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
2*kxsin(θ')-100 / 10.2

3. The Attempt at a Solution :

What I've done so far...
x component of d = d cos(θ) = 1.11 cos(31)= 0.95 (.9514557038)

y component of d = d sin(θ) = 1.11 sin(31) = 0.57 (.5716922632)

added 10 cm to y component = 0.67 (.6716922632)

θ' =tan-1( 0.67 / 0.95)

θ' = 35.19° (35.22066645)

d' = sqrt((x component of d)^2 + (New y component of d) ^2 )

d' = sqrt(0.95^2 + 0.67^2) (1.164662377)

d’ = 1.164662377

x= d' - Unstretched length of spring

x = 1.164662377 - 1.02

x = 0.1446623771 m

exact values (for those wondering):
< = 35.22069645

d' = 1.164662377

mg = 100N

m = 100/g

I used 2 k * x sin(θ') = 100/g * a

And plugged in 2*1060*0.1446623771*sin(35.22069645)*9.8/100 and got 17.33356261 m/s^2 ;and it's still incorrect...

I also tried another equation
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
a = 2*1060*0.1446623771*sin(35.22069645) - 100 / 100/9.8
a = 7.533575457 m/s^2
^(still incorrect)

currently at 71/100 for attempts...

So far, I've slumped 3 tutors on Cheggs, math majors, physics major, you name it. And yes I have tried contacting the teacher for the past few days, no replies back yet. This is starting to be very frustrating.

god bless

p.s.; this is how I got part a:
2*1060*sin(31) = 1091.880719
100/1091.880719 = .0915850956
1.11-.0915850956 = 1.018414904
I rounded it to 1.02 and got it right on webassign (not sure if it made a difference)

 

[haruspex]

Your first two attempts above are clearly wrong in that you forgot to subtract mg. The last two are much better, but I still get a slightly higher number than the last, by about 4%. You need to be particularly accurate when taking a difference of two numbers that are fairly close together, such as happens when you find the new extension. Try keeping 5 digits throughout.

Usually it's better to keep everything algebraic until the very end, only plugging in numbers then. Sometimes this circumvents those 'small difference' steps, other times it suggests an approximation method that's better. But in the present case, the final algebraic expression is pretty messy

 

[alex740rider]

Your first two attempts above are clearly wrong in that you forgot to subtract mg. The last two are much better, but I still get a slightly higher number than the last, by about 4%. You need to be particularly accurate when taking a difference of two numbers that are fairly close together, such as happens when you find the new extension. Try keeping 5 digits throughout.

Usually it's better to keep everything algebraic until the very end, only plugging in numbers then. Sometimes this circumvents those 'small difference' steps, other times it suggests an approximation method that's better. But in the present case, the final algebraic expression is pretty messy

The numbers for the x and y components or the numbers for x in the equation?

 

[SteamKing]

I've been stuck on part b for several hours now (feels like the whole day).

1. Homework Statement :

2. Homework Equations :
2 k * x sin(θ') = 100/g * a
sum(Fy) = F net
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
2*kxsin(θ')-100 / 10.2

3. The Attempt at a Solution :

What I've done so far...
x component of d = d cos(θ) = 1.11 cos(31)= 0.95

y component of d = d sin(θ) = 1.11 sin(31) = 0.57

added 10 cm to y component = 0.67

θ' =tan-1( 0.67 / 0.95)

θ' = 35.19°

d' = sqrt((x component of d)^2 + (New y component of d) ^2 )

d' = sqrt(0.95^2 + 0.67^2)

d’ = 1.162 m

x= d' - Unstretched length of spring

x = 1.162 - 1.02

x = 0.1425 m

mg = 100N

m = 100/g

I agree with your calculations to this point.

I used 2 k * x sin(θ') = 100/g * a

And plugged in 2*1060*.1425*sin(35.19)*9.8/100 and got 17.06 m/s^2 ;and it's still wrong...

As the acceleration on the weight is produced by the net force, you should draw a free body diagram of the weight and the forces acting on it.

another attempt (with more precise numbers) and got 17.33 m/s^2. (no luck)

I also tried
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
a = 7.2644 m/s^2 (still incorrect)

It's not clear what value of 'angle' is being used here.

another attempt (with more precise numbers) and got a = 7.53 m/s^2 (still no luck)

currently at 71/100 for attempts...

So far, I've slumped 3 tutors on Cheggs, math majors, physics major, you name it. And yes I have tried contacting the teacher for the past few days, no replies back yet. This is starting to be very frustrating.

god bless

You're getting close. Don't give up!

 

[Student100]

I've been stuck on part b for several hours now (feels like the whole day).

1. Homework Statement :

2. Homework Equations :
2 k * x sin(θ') = 100/g * a
sum(Fy) = F net
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
2*kxsin(θ')-100 / 10.2

3. The Attempt at a Solution :

What I've done so far...
x component of d = d cos(θ) = 1.11 cos(31)= 0.95

y component of d = d sin(θ) = 1.11 sin(31) = 0.57

added 10 cm to y component = 0.67

θ' =tan-1( 0.67 / 0.95)

θ' = 35.19°

d' = sqrt((x component of d)^2 + (New y component of d) ^2 )

d' = sqrt(0.95^2 + 0.67^2)

d’ = 1.162 m

x= d' - Unstretched length of spring

x = 1.162 - 1.02

x = 0.1425 m

mg = 100N

m = 100/g

I used 2 k * x sin(θ') = 100/g * a

And plugged in 2*1060*.1425*sin(35.19)*9.8/100 and got 17.06 m/s^2 ;and it's still wrong...

another attempt (with more precise numbers) and got 17.33 m/s^2. (no luck)

I also tried
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
a = 7.2644 m/s^2 (still incorrect)

another attempt (with more precise numbers) and got a = 7.53 m/s^2 (still no luck)

currently at 71/100 for attempts...

So far, I've slumped 3 tutors on Cheggs, math majors, physics major, you name it. And yes I have tried contacting the teacher for the past few days, no replies back yet. This is starting to be very frustrating.

god bless

Your work is kind difficult to follow, the linear restoring force of the spring system is $$\vec{F}=-k\vec{x}$$

In your problem we're concerned about the displacement from equilibrium. Obviously to find the instantaneous acceleration we're going to want to look at $$\sum\vec{F}=m\vec{a}$$
Are you looking at the sum of the forces in the direction of acceleration? What're the forces in the y direction upon release?

 

[alex740rider]

Are you looking at the sum of the forces in the direction of acceleration? What're the forces in the y direction upon release?

I have the 2 tension forces upwards and gravity downwards

 

[haruspex]

The numbers for the x and y components or the numbers for x in the equation?

In respect of keeping at least 5 digits? For everything on the path to finding the new extension (so includes finding the relaxed length).
If you keep five, you would normally have four reliable at the end. But the extensions are only about 10% of the relaxed length, so when you take one of those length differences you lose a digit of precision.

 

[alex740rider]

In respect of keeping at least 5 digits? For everything on the path to finding the new extension (so includes finding the relaxed length).
If you keep five, you would normally have four reliable at the end. But the extensions are only about 10% of the relaxed length, so when you take one of those length differences you lose a digit of precision.

also what equation did you use for it to be 4% higher?

 

[ehild]

Keep 4-5 decimals during the calculation steps, including the unstretched length. Calculate x with the more precise value of it. As the lengths are close, you loose one significant digit during subtraction.

 

[alex740rider]

I have updated all the numbers and kept all of the decimals. Can someone tell me what I'm doing wrong or if there's another equation that I can use?

 

[ehild]

We do not see your calculations.

 

[Student100]

We do not see your calculations.

Looks as though he posted it in the original post:

I also tried another equation
Ma = 2kxsin(angle) - 100
a = (2kxsin(angle) - 100) / (100 / 9.8)
a = 2*1060*0.1446623771*sin(35.22069645) - 100 / 100/9.8
a = 7.533575457 m/s^2
^(still incorrect)

 

[ehild]

Looks as though he posted it in the original post:

The original length is not 1.02 m with 4 digit accuracy.

 

[ehild]

I have updated all the numbers and kept all of the decimals. Can someone tell me what I'm doing wrong or if there's another equation that I can use?

Did you updated the original length too?

 

[alex740rider]

Did you updated the original length too?

2*1060*sin(31) = 1091.880719
100/1091.880719 = .0915850956
1.11-.0915850956=
1.018414904

 

[alex740rider]

Did you updated the original length too?

WOW i guess it did made a difference

 

[ehild]

WOW i guess it did made a difference

You see at last? Never round down too much during calculations.

 

[alex740rider]

Solve it, thanks to ehild! You had to use all the numbers in part a to solve it correctly (instead of just 1.02)
Then plug it back into a = (2kxsin(angle) - 100) / (100 / 9.8)
which you will get 7.7235 m/s^2
Thank you to everyone that tried to help!

Again big shout out to ehild!

 

[ehild]

Solve it, thanks to ehild! You had to use all the numbers in part a to solve it correctly (instead of just 1.02)
Then plug it back into a = (2kxsin(angle) - 100) / (100 / 9.8)
which you will get 7.7235 m/s^2

That is exactly the same result I got.
You will remember the importance of significant digits forever I hope, But you need to round off to the desired accuracy at the end. It is 3 digits in this problem, so you have to give the acceleration as 7.72 m/s².

 

[haruspex]

WOW i guess it did made a difference

I guess you did not read this properly:

In respect of keeping at least 5 digits? For everything on the path to finding the new extension (so includes finding the relaxed length).