Work and Energy 1. Work Energy Work done by a constant force (scalar product) Work done by a varying force (scalar product & integrals) 2. Kinetic Energy Chapter 6: Work and Energy Work-Energy Theorem
Jan 05, 2016
Work and Energy
1. Work Energy Work done by a constant force
(scalar product)
Work done by a varying force
(scalar product & integrals)
2. Kinetic Energy
Chapter 6: Work and Energy
Work-Energy Theorem
Work and Energy
Work and Energy
Work by a Baseball Pitcher
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 Done by a Constant Force (I)
Work (W) How effective is the force in moving a
body ?
W [Joule] = ( F cos ) d
Both magnitude (F) and directions () must be taken into account.
Work and Energy
Work Done bya Constant Force (II)
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
Example 1A
A 50.0-kg crate is pulled 40.0 m by a
constant force exerted (FP = 100 N and
= 37.0o) by a person. A friction force Ff =50.0 N is exerted to the crate. Determinethe work done by each force acting on thecrate.
Work and Energy
Example 1A (cont’d)
WP = FP d cos ( 37o )
Wf = Ff d cos ( 180o )
Wg = m g d cos ( 90o )
WN = FN d cos ( 90o )
180o
90o
d
F.B.D.
Work and Energy
Example 1A (cont’d)
WP = 3195 [J]
Wf = -2000 [J] (< 0)
Wg = 0 [J]
WN = 0 [J]
180o
Work and Energy
Example 1A (cont’d)
Wnet = Wi
= 1195 [J] (> 0)The body’s speed
increases.
Work and Energy
Work-Energy Theorem
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
Example 2
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
Example 2 (cont’d)
(1) Wnet = F d(a) cos 180o = - F d(a) = 0 – m v(a)
2 / 2 - F x (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 - F x (? m) = - m (33.3 m/s)2 / 2
(3) F & m are common. Thus, ? = 80.0 m
Work and Energy
Forces
Forces on a hammerhead
Work and Energy
Work and Energy
Work and Energy
Spring Force (Hooke’s Law)
FS(x) = - k x
FPFS
Natural Length x > 0
x < 0
Spring Force(Restoring Force):The spring exerts its force in thedirection opposite the displacement.
Work and Energy
Work Done to Stretch a Spring
x2
W = FP(x) dxx1
FS(x) = - k x
W
Natural Length
FPFS
Work and Energy
Work and Energy
lb
W = F|| dl la
Work Done bya Varying Force
l 0
Work and Energy
Example 1A
A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the person do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the person do ?
Work and Energy
(a) Find the spring constant k k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m(b) Then, the work done by the person is WP = (1/2) k xmax
2 = 1.1 J
(c) x2 = 0.030 mWP = FP(x) d x = 1.1 J x1 = 0
Example 1A (cont’d)
Work and Energy
Example 1B
A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the spring do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the spring do ?
Work and Energy
(a) Find the spring constant k k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m(b) Then, the work done by the spring is
(c) x2 = -0.030 m WS = -1.1 J
x2 = -0.030 mWS = FS(x) d x = -1.1 J x1 = 0
Example 1B (cont’d)
Work and Energy
Example 2
A 1.50-kg block is pushed against a spring(k = 250 N/m), compressing it 0.200 m, andreleased. What will be the speed of theblock when it separates from the spring at
x = 0? Assume k =0.300.
(i) F.B.D. first !(ii) x < 0
FS = - k x
Work and Energy
(a) The work done by the spring is
(b) Wf = - kFN (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) m v2 – 0 v = 2.34 m/s
x2 = 0 mWS = FS(x) d x = +5.00 J x1 = -0.200 m
Example 2 (cont’d)
Work and Energy