Chapter 6: Work and Energy. Work Energy Work done by a constant force (scalar product) Work done by a varying force (scalar product & integrals) Kinetic Energy. Work-Energy Theorem. Work by a Baseball Pitcher. A baseball pitcher is doing work on
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Work done by a constant force
(scalar product)
Work done by a varying force
(scalar product & integrals)
Work-Energy Theorem
Work and Energy
Work and Energy
A baseball pitcher is doing work on
the ball as he exerts the force over
a displacement.
v1 = 0
v2 = 44 m/s
Work and Energy
Work(W)
How effective is the force in moving a body ?
Both magnitude (F) and directions (q ) must be taken into account.
W[Joule] = ( F cos q ) d
Work and Energy
Example:Work done on the bag by the person..
Special case: W = 0 J
a) WP = FP d cos ( 90o )
b) Wg = m g d cos ( 90o )
Nothing to do with the motion
Work and Energy
A 50.0-kg crate is pulled 40.0 m by a
constant force exerted (FP = 100 N and
q = 37.0o) by a person. A friction force Ff =
50.0 N is exerted to the crate. Determine
the work done by each force acting on the
crate.
Work and Energy
F.B.D.
WP = FP d cos ( 37o )
Wf = Ff d cos ( 180o)
Wg = m g d cos ( 90o)
WN = FN d cos ( 90o)
180o
d
90o
Work and Energy
WP = 3195 [J]
Wf = -2000 [J] (< 0)
Wg = 0 [J]
WN = 0 [J]
180o
Work and Energy
Wnet = SWi
= 1195[J] (> 0)
Work and Energy
Wnet= Fnet d = ( m a ) d
= m [ (v2 2 – v1 2 ) / 2d ] d
= (1/2) m v2 2 – (1/2) m v1 2
= K2 – K1
Work and Energy
A car traveling 60.0 km/h to can brake to
a stop within a distance of 20.0 m. If the car
is going twice as fast, 120 km/h, what is its
stopping distance ?
(a)
(b)
Work and Energy
(1)Wnet = F d(a) cos 180o
= - F d(a) = 0 – m v(a)2 / 2
- Fx (20.0 m) = -m (16.7 m/s)2 / 2
(2) Wnet = F d(b) cos 180o
= - F d(b) = 0 – m v(b)2 / 2
- Fx (? m) = -m (33.3 m/s)2 / 2
(3)F & m are common. Thus, ? = 80.0 m
Work and Energy
Forces on a hammerhead
Forces
Work and Energy
Work and Energy
Work and Energy
FS
Spring Force
(Restoring Force):
The spring exerts its force in the
direction opposite
the displacement.
FP
x > 0
Natural Length
x < 0
FS(x) = - k x
Work and Energy
FS
FP
FS(x) = - k x
Natural Length
x2
W = FP(x) dx
x1
W
Work and Energy
Work and Energy
lb
W = F||dl
la
Dl 0
Work and Energy
A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the person
do ? If, instead, the
person compresses
the spring 3.0 cm,
how much work
does the person do ?
Work and Energy
x2 = 0.030 m
WP = FP(x) d x = 1.1 J
x1 = 0
Work and Energy
A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the spring
do ? If, instead, the
person compresses
the spring 3.0 cm,
how much work
does the spring do ?
Work and Energy
x2 = -0.030 m
WS = FS(x) d x = -1.1 J
x1 = 0
Work and Energy
A 1.50-kg block is pushed against a spring
(k = 250 N/m), compressing it 0.200 m, and
released. What will be the speed of the
block when it separates from the spring at
x = 0? Assume mk =
0.300.
FS = - k x
(i) F.B.D. first !
(ii) x < 0
Work and Energy
(a) The work done by the spring is
(b) Wf = - mkFN (x2 – x1) = -4.41 (0 + 0.200)
(c) Wnet = WS+ Wf = 5.00 - 4.41 x 0.200
(d) Work-Energy Theorem: Wnet=K2 – K1
4.12 = (1/2) mv2 – 0
v = 2.34 m/s
x2 = 0 m
WS = FS(x) d x = +5.00 J
x1 = -0.200 m
Work and Energy
Work and Energy