Problem F14-5 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Work and Energy

Principal of work and energy.

When s = 0.6 m, the spring is unstretched and the 10-kg block has a speed of 5 m/s down the smooth plane. Determine the distance s when the block stops.

[kevincamarillo]

the algebra doesnt make sense for teh quadratic equation

s^2: -100
s: 100-49.05 = 50.95
constant: 125-60+36+29.43 = 130.43

tell me if im wrong please

[E]

when i do the simple math i get 100s2+50.95s+130.43

how do you get those numbers the 100s2+ 198.1s-42.14

[Infynte-ky]

I thought since the plane the block is sliding is smooth therefore, there is no friction. So then why are we figuring out T_1 if it is supposed to be 0?

[hanzedwardmaningo]

Where did you get the 198.15 and 42.14 after you simplyfying?

[rickjljr]

I realize this is old and not much interest, but does anyone have a published answer key for this problem? As a challenge I was trying to solve the ODE
d^2x/dt^2 = -k/m*(x-0.6) + 149/m with x0 = 0.6 and v0 = 5.
YouTube has 4 videos working this problem with 4 (maybe 6) different answers. After a lot of work I came out with s = 2.6889 but still not sure.

[mahmoud]

hallo
thank you for your solution. only I have a comment on Spring's energy because you got it wrong
Us=0.5K(s2-0.6)^2 ... Not Us=0.5K((s2)^2-(0.6)^2 )
the sping was unstretched when s=0.6 so x1=0, then the spring was stretched so that x2=S2-0.6
Thanks again for your great job

[jimfoug]

I tried solving this using F=ma and net force but for some reason i am getting the wrong answer.
Net force (-30degrees)= 100 + sin(30)mg -200s
ma = 149.05 - 200s
10v dv = (149.05- 200s)ds
then i integrated with v going from 5 to 0 and and s going from .6 to s.
My answer was 1.873 meters. Any idea why it was wrong? thank you

[victorhu]

I don't think you are doing it right, because there's no work done in the spring when it's equal to 0.6, it's supposed to be (S2-0.6)square

[iskendersefaylmaz]

I think your answer is not correct. While calculating the spring energy, the first length should be taken as 0 not 0.6

[abdullahelgamal]

Its + cos(30) not -sin(30) because its on a decline so when we resolve the wight its going to be to the right and down we need the force to the right cause its gonna push with the force so +10x9.81cos(30)(S-0.6) the rest is correct

[HashemAljifri]

Your answer is wrong. The correct answer is s= 1.873 + 0.6 = 2.473 meters.