Block A (mass = ma) is attached to the left end of an ideal (massless) spring wi
ID: 1481610 • Letter: B
Question
Block A (mass = ma) is attached to the left end of an ideal (massless) spring with stiffness value of k.
Block B (mass = mb) is not attached to anything. Both masses are supported by a level surface.
That surface is frictionless in the region where block A can travel, but it offers friction (uk, us) in the region where block B can travel. There is no air drag. Initially, block B is at rest, and block A is moving to the right at a known speed vA (with the spring relaxed). Then the spring's right end collides with block B.
Summary of known values: mA, k, mB, uk, us, vA, g.
Show (with full diagrams and/or equation, as necessary) how you could calculate whether block B slips before block A stops.
Explanation / Answer
if A stops then
using energy conservation
0.5 * k * X ^2 = 0.5 * mA * vA ^2
X = sqrt(mA * vA^2 /k)
now this compression in spring will apply force on B
kX = force spring exerted on B
us * mB * g =minimum force that is required to move B
if Kx > us * mB * g
then it means B slips before A comes to stop