Physic 231 Lecture 10
•
•
Main points of last lecture:
Frictional forces:
– kinetic friction:
f k = µk N
•
•
•
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More examples.
W = F∆s cos(θ ) = Fx ∆x v x
Kinetic energy:
KE =
– static friction
f s < µs N
Main points of today’s lecture:
Work:
•
1 2
mv
2
F θ
v
∆s
Work-energy theorem:
Wtotal = KE f − KE0
•
Potential energy:
∆PE = PE f − PE0 = mg ( y f − y0 )
Work
•
Physicists have a precise definition of work done by a force on an
object. It is computed from the displacement of the object while being
acted on by the force.
v
F
∆x
θ
v
∆s
W = F∆s cos(θ ) = F∆x
•
Note: work is only done if there is a displacement in the direction of the force.
– if θ = 0o, W = F∆s
– if θ = 90o, W = 0.
– if θ = 180o, W = -F∆s
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The units for work are N⋅m = J (Joules). These are the units we will use in this
chapter. Other units are cal (calorie). 1 cal = 4.186J. Calorie (Food calories) is
another unit. 1 Calorie = 1000 cal.
Work is not the same thing as effort.
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What work is being done by me?
•
Indicate with your response unit whether:
– a) positive work is being done
– b) no work is being done
– c) negative work is being done
What work is being done by gravity?
•
Indicate with your response unit whether:
– a) positive work is being done
– b) no work is being done
– c) negative work is being done
What work is being done by me?
•
Indicate with your response unit whether:
– a) positive work is being done
– b) no work is being done
– c) negative work is being done
What work is being done by friction?
•
Indicate with your response unit whether:
– a) positive work is being done
– b) no work is being done
– c) negative work is being done
What work is being done by the string?
–
–
–
–
a) positive work
b) negative work
c) zero work
d) cannot be determined
Example
•
A workman lifts a 4 kg brick 1.5 meter vertically. a) What is the work
done by the workman? b) What is the work done by gravity? c) If the
workman lowers it to the ground, what work does he do in lowering
the brick? d) What would be his total work? Assume all motions are at
constant velocity.
m
Wy = − mg = −39.2 N
4 kg
Fman , y = − W = 39.2 N
∆y parts
a&b
1.5 m
a) Wman ,up = Fy ∆y = (39.2 )(1.5)J = 59J
∆y parts
a&b
-1.5 m
b) Wgrav,up = − mg∆y = −59J
v
Fman
v
W
c) Wman ,down = Fy ∆y = (39.2 )(− 1.5)J = −59J
d ) Wman , total = Wman ,up + Wman ,down = 0
Work-energy theorem, kinetic energy
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Let us consider the case of a constant force F in the x direction. Then
W = F∆x
By Newton’s second law, F=max. Then
•
•
W = ma x ∆x
But
2a x ∆x = v x2 − v x2,0 ⇒ W = ma x ∆x =
•
1 2 1 2
mv x − mv x ,0
2
2
There is no force in the y direction, ay=0, vy=vy,0 and 1/2mvy 2= 1/2mvy,0 2.
Thus
1
1
⇒W=
•
We define,
mv 2 − mv02
2
2
1 2
KE = mv
2
to be the kinetic energy. In terms of KE,
W = KE − KE0
Quiz
•
A 1000 kg car is moving at 30 m/s, when the driver applies the brakes.
After the brakes are applied, the car rolls to a stop in 100 m. How
much work is done by the brakes in bringing the car to a stop?
– a) 30000 N
– b) -15000 J
1
2
KE
=
mv
0
0
– c) -4.5x105 J
2
– d) 1.5x105 J
1
2
KE f =
2
mv = 0
1
W = KE f − KE 0 = − mv 02
2
1
2
W = − ⋅ (1000 )(30 ) J = −4.5x105 J
2
Conceptual problem
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A cart on an air track is moving at 0.5 m/s when the air is suddenly
turned off.The cart comes to rest after traveling 1 m.The experiment is
repeated, but now the cart is moving at 1 m/s when the air is turned off.
How far does the cart travel before coming to rest?
–
–
–
–
–
1
KE 0 = mv 02
2
1
KE f = mv 2 = 0
2
a) 1 m
b) 2 m
c) 3 m
d. 4 m
1
2
=
−
=
−
W
KE
KE
mv
f
0
0
e) impossible to determine
2
1
W = −f k ∆x = − mv 02
2
⇒
KE try 2
KE try1
2
f k ∆x try 2 ∆x try 2
 v2 
=   = 4 =
=
f k ∆x try1 ∆x try1
 v2 
Example
•
A 1 kg block is thrown downward with an initial velocity of 10 m/s.
What is the kinetic energy and speed of the block when it strikes the
ground 10 m below. Would it matter whether the block were thrown
horizontally as in trajectory 1 or at other angles such as trajectories 2
or 3?
KE f − KE0 = Work g = ( − mg )( −10m)
KE f = KE0 + ( mg )(10m)
10 m/s
10 m
1
(1kg )(10m / s ) 2 + (1kg )(9.8 N )(10m) = 148 J
2
2(148J)
1 2
mv = 148 J ; v =
= 17.2m / s
1 kg
2
KE f =
Fig. 5.15, p. 127
Slide 16
The final speed and kinetic energy are
independent of the initial angle.