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
Sorry, No idea. LOL.
sigh...
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.
so if E is 192
that's using 1/2 (10.0 - 4.0) (8.0)^2
how do i find compression distance?
E=1/2kx^2
192=1/2Kx^2
so 192 = 1/2 (4000) and solve for x^2?
yes
okay...i get .3098386677 which isn't an option
you did consider momentum transfer though right?
E=1/2mv^2 is the total energy of the blocks.... there is completely elastic collision that happens...
find the final speed of the 2 blocks by: m1v1=(m1+m2)V
after you find that THEN use the E=1/2mV^2 equation
(m1+m2)V is 112
then i plug that into E = 1/2 (112)(8.0)^2
i get 3584
no the V changes..... find V in the momentum transfer equation you know the rest
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