Spring constants! power

[Jessicaelleig]

Two identical massless springs of constant k = 200 N/m are fixed at opposite ends of a level track, as shown in Figure P5.62. A 5.90 kg block is pressed against the left spring, compressing it by 0.150 m. The block (initially at rest) is then released, as shown in Figure P5.62a. The entire track is frictionless except for the section between A and B.

Figure P5.62.

Given that the coefficient of kinetic friction between block and track along AB is µk = 0.0800, and given that the length AB is 0.238 m,
(a) determine the maximum compression of the spring on the right (see Fig. P5.62b).

(b) Determine where the block eventually comes to rest, as measured from A (see Fig. P5.62c).


If anyone can provide some guidance, I would appreciate it.
Thank you

[LowlyPion]

Two identical massless springs of constant k = 200 N/m are fixed at opposite ends of a level track, as shown in Figure P5.62. A 5.90 kg block is pressed against the left spring, compressing it by 0.150 m. The block (initially at rest) is then released, as shown in Figure P5.62a. The entire track is frictionless except for the section between A and B.

Figure P5.62.

Given that the coefficient of kinetic friction between block and track along AB is µk = 0.0800, and given that the length AB is 0.238 m,
(a) determine the maximum compression of the spring on the right (see Fig. P5.62b).

(b) Determine where the block eventually comes to rest, as measured from A (see Fig. P5.62c).


If anyone can provide some guidance, I would appreciate it.
Thank you

Since you only have the one area that has friction - where energy can be robbed from the block - then what you are interested is how much 1 pass over the patch of friction will take away.

W = F*d

so your work each pass removed is u*m*g*d = .08 * 5.9 * 9.8 * .238

The work for the Spring is kx2/2 so knowing how much work goes to KE then gets subtracted each pass over the patch you can figure the rest.

[Jessicaelleig]

thanks! anything else? im so confused

[LowlyPion]

thanks! anything else? im so confused

Any more and I'd be giving you the answers wouldn't I?

And gosh darn it I can't take your tests for you. So what good what that be?

Just draw a diagram and figure it out.