A 5.430 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 5.430 kg block of wood rests on a steel desk. The coefficient of static friction between the block arid the desk is Mu=0.555 and the coefficient of kinetic friction is Muk=0.255. At time t = 0, a force F = 18.2 N Is applied horizontally to the block. State the force of friction applied to the block by the table at the following times: t = 0 t > 0 Number Number N N Consider the same situation, but this time the external force F is 367 N Again state the force of friction acting on the block at the following times: t = 0 t > 0 Number Number N N

Explanation / Answer

a) Fstaticmax = us N = 0.555*5.43*9.81= 29.56 N

Not enough to overcome static so static will just cancel force

so t = 0, friction = 18.2

t>0, friction = 18.2 N

b) now enough to overcome friction

so F kinetic = 0.255*5.43*9.81 = 13.6 N

so t = 0 friction = 29.56 N

t>0 friction = 13.6 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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