Is the textbook wrong or am I?

  • Thread starter Thread starter drewdiddy
  • Start date Start date
  • Tags Tags
    Textbook
AI Thread Summary
The discussion revolves around a physics problem involving a vertical spring and the speed it can impart to a ball when released. The initial calculations yield a velocity of 7.68 m/s, while the textbook states it should be 7.47 m/s. Participants clarify that the conservation of energy equation must account for gravitational potential energy, leading to the conclusion that the textbook's answer is correct. One participant acknowledges the error in their terminology regarding vertical distance. Ultimately, the consensus is that the textbook's calculation is accurate.
drewdiddy
Messages
13
Reaction score
0

Homework Statement



A vertical spring (ignore its mass), whose spring constant is 875 N/m is attached to a table and is compressed down by .160 m. (a) What upward speed can it give to a .380 kg ball when released?


Homework Equations



Conservation of Energy using 1/2 k x^2 for Uspring.

The Attempt at a Solution



I get 7.68 m/s for the velocity and the book gets 7.47 m/s. Wanted to see who was right and if I'm doing something wrong.
 
Physics news on Phys.org
drewdiddy said:

Homework Statement



A vertical spring (ignore its mass), whose spring constant is 875 N/m is attached to a table and is compressed down by .160 m. (a) What upward speed can it give to a .380 kg ball when released?


Homework Equations



Conservation of Energy using 1/2 k x^2 for Uspring.

The Attempt at a Solution



I get 7.68 m/s for the velocity and the book gets 7.47 m/s. Wanted to see who was right and if I'm doing something wrong.

"Upward" also means against gravity. You should also figure as an adjustment the m*g*h over the displacement of the acceleration.
 
So you're saying you got the book's answer?

I used the conservation of energy subbing values for spring and taking into account y=0 when crossing the original spring length. I'm quite sure I have the right answer and the book's is wrong but I just want to verify.
 
drewdiddy said:
So you're saying you got the book's answer?

I used the conservation of energy subbing values for spring and taking into account y=0 when crossing the original spring length. I'm quite sure I have the right answer and the book's is wrong but I just want to verify.

I'm just saying that

mv2/2 = kx2/2 - m*g*x
 
Last edited:
You are wrong, your book is right, Pion is right: (except I would not use both "h" and "x", there is only one vertical distance in the problem)
 
borgwal said:
You are wrong, your book is right, Pion is right: (except I would not use both "h" and "x", there is only one vertical distance in the problem)

Thanks for the catch. Of course h and x are the same.

I edited the previous post to be correct now.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Projectile Motion : Work and Energy
Replies
4
Views
1K
A rocket on a spring, related to potential/kinetic energy
Replies
3
Views
2K
Question about springs and rotational/translational energy
Replies
4
Views
1K
How High Does Cheese Rise on a Released Spring?
Replies
1
Views
1K
Find elastic energy in the compressed spring.
Replies
12
Views
3K
Solve Masses on a Spring: Compression Distance D vs d
Replies
2
Views
2K
A block of mass m is dropped onto the top of a vertical spring....
Replies
8
Views
5K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K
Conservation of Energy: Spring PE: Toy gun on spring
Replies
4
Views
2K
Determine expressions for the following in terms of M, X, D, h and g
Replies
30
Views
1K

Hot Threads

Recent Insights

Back
Top