ame: Third Test, Page 3 of 5 April 4, 2018 3. The track shown has only a small r
ID: 2035737 • Letter: A
Question
ame: Third Test, Page 3 of 5 April 4, 2018 3. The track shown has only a small rough section, of lgth0.600m that offers friction between the track and the box with a coefficient of kinetic friction,H- 0.600. The track is otherwise smooth. After the rough patch, the track ends on a spring with a force constant k, 1666. /m lenght A box of mass m 93.75kg, initially at rest, is released from a height h 2.400m (I) (5 points) Draw the forces acting on the box as it moves through the rough patch, make a free body diagram for the box and find the value of the friction force, f (do not confuse with (II) (15 points) Find the speed of the box after leaving the rough patch.Explanation / Answer
part I.
forces acting on the body area:
weight of the body=93.75*9.8=918.75 N, in vertically downward direction
normal force, in vertically upward direction=weight of the body=918.75 N
friction force=friction coefficient*normal force
=551.25 N
it is acting in direction opposite to the motion , hence to the left side.
free body diagram:
part II.
before the mass enters the rough patch:
let speed of the mass be m kg.
using conservation of energy principle:
initial potential energy+initial kinetic energy=final potential energy+final kinetic energy
==>mass*g*initial height+0.5*mass*initial speed^2=mass*g*final height+0.5*mass*final speed^2
==>93.75*9.8*2.4+0.5*93.75*0=93.75*9.8*0+0.5*93.75*v^2
==>v=6.8586 m/s
after the mass has passed through the rough patch:
let speed after it has gone through the rough patch is v m/s.
using work-energy principle:
kinetic energy before entering the rough patch -work done against friction=kinetic energy after the rough patch
==>0.5*93.75*6.8586^2-friction force*distance=0.5*93.75*v^2
==>2205-551.25*0.6=0.5*93.75*v^2
==>v=6.3233 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.