英文版物理功和能

Calculate the work by the force when the particle moves from x1=2(m) to x2=3(m) in the following cases: 1. The trajectory is a parabola
Example
x 4y
2
2. The trajectory is a straight line 4 y x 6
Work is a scalar, no direction. π positive work; 0 d W 0 ,
π 2 π 2 π 2 dW 0 , d W 0,
Constant Force (including its direction)
A
rA
o
s
(2) Work done by a varying force: When a particle moves from A to B along a curve path, the total work ds done by the force equals the integral of F from A to B: A
7-1 Work and Power (P147-)
1. Work
研究力和运动的空间过程关系。
The work done by a force F in displacing a s body s as the scalar product of F and s.
W F r cos F r
1
2
O
3 X
A1
( Fx dx Fy dy ) 2 ydx 4dy
x1 y1 2 94 1
x2
y2
( x / 2)dx 4dy 10 . 8 J
2
Work depends on path.
A2
x2 , y 2 x1 , y1 3
( Fx dx Fy dy) 2 ydx 4dy
L1 m1
f12 Force on m1 by m2
o
L2
dA f12 dr12 f 21 dr21
dA f12 dr12 f 21 dr21
We have used this conclusion before Work by pair of normal forces
W1 W2 W3 Fra bibliotekThe work done by several forces = algebraic sum of the work done by the individual force. (4) Special expression in scalar product: F Fx i Fy j Fz k , and dr dxi dy j dzk .
Y x2 4 y
4y x 6
2.25
1
2
O
3
X
5
Solution:
Fx dx Fy dy Fz dz 4 y x 6
b a
A
B
A
F dr
2.25
Y x2 4 y
F 2 yi 4 j ( N )
x2 , y 2 x1 , y1 3
zero work; negative work.
Explanations:
s rB rA —displacement from A to B
(1) If the force is constant in the whole path, B A W F dr F (rB rA ) F s
f12 ,
f12
r1
f21
f 21
m2 r2
d r12 Displacement of m1 w.r.t to m2
f12 dr1 f12 dr2 f12 (dr1 dr2 ) f12 dr12
f12 f 21
x1 y1 94 1
x2
y2
( x 6) / 2dx 4dy 21 .25 J
2
Work of a pair of forces Interaction force between m1 and m2
dA dA1 dA2 f12 dr1 f 21 dr2
W dW
A B B A
rB
B
F
B
F
F dS
Work is dependant on the reference frame.
(3) Work done by several forces:
If F F1 F2 F3 F4 W F d r F1 d r F2 dr F3 dr
W
B A xi
B F dr ( Fx dx F y dy Fz dz )
A

xf
F x dx
yf yi
F y dy Fz dz
zi
zf
(5) Geometrical representation of work: F The work done by a force equals the area under the F versus x curve —— 变力曲线与 x1 dx 位移轴在两点x1, x2 之间所包围的面积.
F F x F
x
x2
x
x
W>0
W<0
W=0
(6) Unit of work: Dimension is: MLT –2 L= M L2 T –2
In SI system, 1 N· m = 1J (joule) = 1kg· m2/s2
The force on the particle is F 2 yi 4 j ( N )