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Question 1,
A
0.23
kg pinecone falls
16.0
m to the ground, where it lands with a speed of
15
m/s.
(a) With what speed would the pinecone have landed if there had been no air resistance?
(b) Did air resistance do positive work, negative work, or zero work on the pinecone?
Question 2:
Work down by Normal force:
W
N
W=N*d*cos(90)
,
angle
between Normal force N (perpendicular to the
surface) and the motion (along the surface) is
90 degrees.
W
N
=0
Work down by gravity:
W
mg
=mg*d*cos
(90),
angle between gravity (downward) and the motion (along
the surface, horizontal) is 90 degrees.
W
mg
=0.
Work down by friction:
W
f
=f
k
*d*cos
(180),
angle between friction and the motion is 180 degrees.
Cos(180)=1 ,
W
f
=

f
k
*d
Friction is kinetic
, f
k
=
µ
k
*N
, N=mg, W=
µ
k
*mg*d*(1)=0.2*9*9.81*5*(1)= 88.2 J
b). Total work is equal to change of kinetic energy:
W
total
=W
friction
=K
f
 K
i
= 88.2 J
So
:
K
f
=K
i
+W
, Final Kinetic energy is equal to initial kinetic Energy plus the work down to it. The work is
negative. K
f
is less than K
i
½ mv
f
2
=½ mv
i
2
+W
friction
, Simplify it by multiply 2/m to both side of the equation,
you get:
v
f
2
= v
i
2
+W
friction
*(2/m)
v
f
2
= 7
2
+(88.2)*(2/9)
v
f
=5.42 m/s
c). From start till become rest, W
total
=K
rest
 K
initial
=0  K
i
=  ½ m v
i
2
=  ½ *9*7
2
= 220 J
This is the short way, only worries about the initial and final kinetic energy.
Since f
k
=
µ
k
*N is a constant force, the acceleration is constant, a=f
k
/m= 
µ
k
*N/m
From the acceleration and
initial and final velocity, you can find out the total distance from the beginning till it stops, using our old friend,
v
f
2
= v
i
2
+2a(
∆
x), (Note that a is negative here.)
.
W
total
=W
friction
= f
k
*
∆
x*(1) = 220 J
This is the longer way.
When you know initial and final velocity, the total work equals to the change of kinetic energy is the best way for
you to find total work. (no need to worry about acceleration at all). This also works for nonstraight motion, for
non constant acceleration, loop, complicated track, etc.
It is always true that: W