Refer to this diagram: Block A (mass = mA) is attached to the left end of an ide

Question

Refer to this diagram: Block A (mass = mA) is attached to the left end of an ideal (massless) spring with a 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 (mu_k, mu_s) 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. Refer to this diagram (not to scale): Show (with full diagrams and/or equations, as necessary) how you could calculate whether block B slips before block A stops.

Explanation / Answer

block A will stop when total initial kinetic energy is converted fully into potential energy of spring.

hence o.5*Ma*v^2=0.5*k*x^2

where x=compression of spring

==>x=sqrt(Ma/k)*v...(1)

now, at highest compression, spring force being exerted on mass B=k*x=k*v*sqrt(Ma/k)=v*sqrt(k*Ma)

then the bock B will slip if highest static friction is less than this force

hence condition for block B slipping will be

mu_s*Mb*g=v*sqrt(k*Ma)....(2)

where mu_s=coefficient of static friction

if mu_s*Mb*g>v*sqrt(Ma*k)

then the block B wont slip

or else the block B will start slipping and the friction force on the block B will change to mu_k*Mb*g

where mu_k=coefficient of kinetic friction

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.