Recent content by bestchemist

Got it! Thank y'all :)

I tried but my answer came out to be something that's not even on the answer choice lol Here is what I did Δk = w 1/2 mv^2 = Fd (1/2)(3)v^2 = 1*8 V= 2.3 m/s

Homework Statement FIGURE 7-6 The force on a 3.00-kg object as a function of position is shown in Fig. 7-6. If an object is moving at 2.50 m/s when it is located at x = 2.00 m, what will its speed be when it reaches x = 8.00 m? ________ A) 2.90 m/s B) 3.30 m/s C)...

okay.. got it

Where can I simplify it? I thought I already simplified it to the most simplest form. Can you give some hint?

(1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2 and solve for v and plug x = L v^2 = 2 (Fo/m)x - 2 uogx+ 2(uog/2L)x^2 v^2 = 2x ((Fo/m)- uog) + (uog/L)x^2 v^2 = 2L((Fo/m)- uog) + (uogL) v= √(2L((Fo/m)- uog) + (uogL)) is it right?

a = (F0 - mg(1-x/L))/m vdx = (Fo/m)dx - uogdx + (uog/L)xdx (1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2 and solve for v and plug x = L I got √(2L(F0m−μ0g))+√μ0gL

find an expression for the block's velocity when it reaches position x=L. Express your answer in terms of the variables L, F0, m, μ0, and appropriate constants.

Homework Statement A block of mass m is at rest at the origin at t=0. It is pushed with constant force F0 from x=0 to x=L across a horizontal surface whose coefficient of kinetic friction is μk=μ0(1−x/L). That is, the coefficient of friction decreases from μ0 at x=0 to zero at x=L...

yayyyyyyyyyyyy... After a few days on this one problem... Finally got it lol Thank you so much!

I think I got it for real this time... lol sooo much confident lol U = 2x^2 -4x +1 Right?

I think I got it... lol U = 2x^2 -6x +2 x = 0.5 U = 2(0.5)^2 -6(0.5)+2 U =-0.5 Right?

okie, so U = -2x^2 -4x +2 right? If x = 0.5, then U = -2(0.5)^2 -4(0.5)+2 U=-0.5

okay... so If I did it right U = 2x^2 -4x-2 and for x >1 I did the calculation and I'm so sure about it lol U = -1 right?

ohhh... yep I see it now... I didn't look at the limit like you said... does the first term is -1/2 then since -2x^2 ]0.5 = -1/2 how about from x -1 to 1 then... how will the function become constant since there are so many variable here