30.0 kg block is resting on a flat horizontal table. On top of this block is res

Question

30.0 kg block is resting on a flat horizontal table. On top of this block is resting a 15.0 kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 320 N/m. The coefficient of kinetic friction between the lower block and the table is 0.625, while the coefficient of static friction between the two blocks is 0.930. A horizontal force F is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed.
At the point where the upper block begins to slip on the lower block determine the following.
(a) the amount by which the spring is compressed
(b) the magnitude of the force F

Explanation / Answer

N=15*g. friction force between the two block. N*0.93=137(N). so that F_spring = 137N. so ?x=137/320=0.43(m). -------- friction force with the table. N=45*g F=N*0.625=275(N). this is larger than F_spring=137 N so no need another force to keep it balance, so F=0

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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