A 4.700 kg block of wood rests on a steel desk. The coefficient of static fricti
ID: 1348665 • Letter: A
Question
A 4.700 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.555 and the coefficient of kinetic friction is k = 0.205. At time t = 0, a force F = 15.7 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
Consider the same situation, but this time the external force F is 31.7 N. Again state the force of friction acting on the block at the following times:
t=0 ? t >0 ?
Explanation / Answer
at t=0
the force of friction =s*N
normal force=mg
friction =0.555*4.7*9.8=25.56 N
at t>0 kinetic friction=k*N
kinetic friction=0.205*4.7*9.8=9.44 N
2) static friction=0.555*4.7*9.8=25.56 N
kinetic friction=0.205*4.7*9.8=9.44 N
friction mainly depends on :
1.area of surface contact
2. nature of surface contact
3. normal force
4. mass of the block
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.