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Mathematics 84 Online
OpenStudy (anonymous):

a block of mass 10.0 kg slides without friction at a speed of 8.0 m/s on a horizontal table surface until it strikes and sticks to a mass of 4.0 kg attached to a horizontal spring (with a spring constant k = 4000 N/m), which in turn is attached to the wall. How far is the spring compressed before the mass comes to rest

OpenStudy (saifoo.khan):

Sorry, No idea. LOL.

OpenStudy (anonymous):

sigh...

OpenStudy (anonymous):

Find the kinetic energy of the block... E=1/2mv^2 this is turned into potential energy in the spring.... you can find he compression distance from that.

OpenStudy (anonymous):

so if E is 192

OpenStudy (anonymous):

that's using 1/2 (10.0 - 4.0) (8.0)^2

OpenStudy (anonymous):

how do i find compression distance?

OpenStudy (anonymous):

E=1/2kx^2

OpenStudy (anonymous):

192=1/2Kx^2

OpenStudy (anonymous):

so 192 = 1/2 (4000) and solve for x^2?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay...i get .3098386677 which isn't an option

OpenStudy (anonymous):

you did consider momentum transfer though right?

OpenStudy (anonymous):

E=1/2mv^2 is the total energy of the blocks.... there is completely elastic collision that happens...

OpenStudy (anonymous):

find the final speed of the 2 blocks by: m1v1=(m1+m2)V

OpenStudy (anonymous):

after you find that THEN use the E=1/2mV^2 equation

OpenStudy (anonymous):

(m1+m2)V is 112

OpenStudy (anonymous):

then i plug that into E = 1/2 (112)(8.0)^2

OpenStudy (anonymous):

i get 3584

OpenStudy (anonymous):

no the V changes..... find V in the momentum transfer equation you know the rest

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