Block falls on Spring Work and Speed

AI Thread Summary
A block weighing 240 grams is dropped onto a vertical spring with a spring constant of 11 N/cm, compressing it 15 cm before stopping. The work done on the block by gravitational force is calculated to be approximately 0.3528 J. For the work done by the spring force, integration yields a result of 12.375 J, though there is uncertainty about its accuracy. The user struggles to determine the block's speed just before impact and questions how to find the height from which it fell. Despite some progress on parts of the problem, the user remains confused about the calculations and seeks clarification.
BlasterV
Messages
38
Reaction score
0
A block of mass 240 grams is dropped onto a relaxed vertical spring that has a spring constant 11 N/cm as shown on the figure. The block becomes attached to the spring and compresses the spring 15 cm before momentarily stopping.

First Conversions:
240g = .24kg, 11 N/cm = 1100 N/m, 15cm = .15m

While the spring is being compressed, what work is done on the block by the gravitational force on it?

.24kg * 9.8 m/s^2 = 2.352 N
2.352 N * .15m = .3528 J -> This is definitley correct

What work is done on the block by the spring force?

Work done over distance via a function so use integration:

W = 0 to .15m ( 1100N/m x dx ) =
W = 0 to .15m ( 1100N/m x^2/2 )
W = (.15m)^2 / 2 * 1100N/m ) = 12.375 J -> No idea if this is right.


What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

This is where I have problems, I have NO idea how to find the speed of the block. I don't know how high it falls from, I tried using 9.8m/s^2 as an accel and VF = 0 in vf^2 = vi^2 + 2ad, but no go, can anyone shed light on this part?


If the speed at impact is doubled, what is the maximum compression in the spring?

No idea on this either, I don't know the answer to part 3 is probably why.


Any help on this would be great!
 
Physics news on Phys.org
I managed to get part 3 and 4, but the Part 2 with 12.375 J is still wrong, It looks perfect to me, did I miss something?
 
Got it all even though no one helps, thanks anyway lol, someone delete this please?
 
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...
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...
Back
Top