Download F ma = Or , 0 , 0 Integrating twice gives the solution for position at

3.4 Newtonian Mechanics Newton’s Second Law of Motion states that when the net force acting on a body is not zero, then the net force is proportional to its acceleration or more precisely F
i
 ma i
Now suppose a rock is tossed upward from the roof of a building. If we assume that the upward direction is positive and that no forces act on the rock other than gravity, then Newton’s second law gives Or
,
0
,
0
Integrating twice gives the solution for position at time t:
1
2
1. A 400-lb object is released from rest 500 feet above the ground and allowed to fall under the
influence of gravity. Assuming that the force in pounds due to air resistance is -10v, and where v
is the velocity of the object, determine the equation of motion of the object. When will the object
hit the ground?
F
i
 ma
i
mg  10v  m
dv
dt
mg  400
m(32)  400
400
 12.5
32
dv
mg  10v  m
dt
m
400  10v  12.5
dv
 0.8v  32
dt
dv
dt
v(0)  0, x(0)  0
The integrating factor is e 
d 0.8t
 e v   32e0.8t
dt 
d 0.8t
0.8 t
 dt e v  dt   32e dt
e0.8t v  40e0.8t  c
v  40  ce 0.8t
0.8 dt
 e0.8t
v(0)  0
0  40  ce0
40  c
v  40  40e 0.8t


x   40  40e 0.8t dt
x  40t  50e 0.8t  c
0  40(0)  50  c
c  50
x(0)  0
x  40t  50e 0.8t  50
500  40t  50e 0.8t  50
0  40t  50e 0.8t  550 Using a calculator t  13.75 seconds
2. An object of mass 8 kg is given an upward initial velocity of 20 m/sec and then allowed to fall
under the influence of gravity. Assume that the force in newtons due to air resistance is -16v.
Determine the equation of motion of the object. If the object is initially 100 m above the ground,
determine when the object will strike the ground.
 Fi  ma
i
dv
dt
mg  8  9.8   78.4
 mg  16v  m
dv
dt
dv
78.4  16v  8
dt
dv
 2v  9.8 v(0)  20, x(0)  100
dt
2 dt
The integrating factor is e   e 2t
 mg  16v  m
d 2t
 e v   9.8e 2t
dt
d 2t
2t
 dt e v  dt   9.8e dt
e 2t v  4.9e 2t  c
v  4.9  ce 2t
v(0)  20
20  4.9  ce0
24.9  c
v  4.9  24.9e 2t


x   4.9  24.9e 2t dt
x  4.9t  12.45e 0.8t  c
100  4.9(0)  12.45  c
c  112.45
x(0)  100
x  4.9t  12.45e 2t  112.45
0   4.9t  12.45e 2t  112.45 Using a calculator
t  22.95 seconds
3. Suppose a small cannonball weighing 16 pounds is shot vertically upwards from ground level
300 /
.The projectile is subject to an air resistance that is
with an initial velocity
proportional to the instantaneous velocity with constant of proportionality
0.0025. What is
the maximum height of the cannon ball?
4. A parachutist whose mass is 100 kg drops from a helicopter hovering 3000 m above the
ground and falls under the influence of gravity. Assume that the force due to air resistance is
proportional to the velocity of the parachutist, with proportionality constant 20 N-sec/m when the
chute is closed and 100 N-sec/m when the chute is open. If the chute does not open until 30 sec
after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the
chute does not open until 1 minute after he leaves the helicopter, after how many seconds will he
hit the ground?
5. When an object slides on a surface, it encounters a resistance force called friction. This force
has a magnitude of  , where  is the coefficient of kinetic friction and N is the magnitude of
normal force that the surface applies to the object. Suppose an object of mass 30 kg is released
from the top of an incline that is inclined 300 to the horizontal. Assume the gravitational force is
constant, air resistance is negligible, and the coefficient of kinetic friction   0.2 . Determine
the equation of motion for the object as it slides down the plane.. If the top of the surface is 5 m
long, what is the velocity of the object when it reaches the bottom?