4. Power(功率) p dw F da=F,(3-8)(请同学自学) d dt The unit of p is lw=1J/s 啦()画
4. Power （功率） The unit of P is 1w = 1J/s . t F V （3-8） F r t W P = = = d d d d （请同学自学）
Example3-1:一物体在x轴上运动,受到力F=5x的作用,求物体 从xn=1运动到xb=2过程中,F所作的功。 解:根据功的定义,有 w= Fdx=L-5xdx 5 15 2
Example 3-1：一物体在x轴上运动，受到力F=-5x的作用，求物体 从 xa = 运动到 xb = 过程中，F所作的功。 解：根据功的定义，有 J x W Fdx xdx b a x x = − = − = = −
Example32 Work done by friction(摩擦力功):a small body of mass m acted upon by an frictional force moves from a to b on a horizontal(水平) table with a coefficient of kinetic friction u. Calculate the work done by frictional force: (1the path is a half of circle of radius (半径)R,(2) the path is the diameter(直径)AB Solution: The element work is A B dW=f6·d=JcO丌d=-1 mg ds (1) The path is arc AB (路径是半圆弧) W1=「fG=-"1ds=- ur marR
Example 3.2 Work done by friction (摩擦力功）：a small body of mass m acted upon by an frictional force moves from A to B on a horizontal（水平） table with a coefficient of kinetic friction . Calculate the work done by frictional force: ⑴the path is a half of circle of radius （半径） R, ⑵ the path is the diameter(直径） AB. k Solution: The element work is W f r f os s mg s d = k d = k c d = −k d W f r mg s k mg R B A AB = k = −k = − d d ⑴ The path is arc AB ⌒ （路径是半圆弧） A A B
(2) The path is diameter AB(路径是直径) B W2=「 u,ng ds=-2 AUk JAB u mgR which means that the work done by frictional force depends on the path B (摩擦力的功取决于路径。) 以后知道这类力是非保守力,没有相关的势能
W f r mg s k mgR B A AB 2 = k d = −k d = −2 which means that the work done by frictional force depends on the path. （摩擦力的功取决于路径。） ⑵ The path is diameter AB￣（路径是直径） A A B 以后知道这类力是非保守力，没有相关的势能
Example33 Work done by a spring(弹性力的功) with a force constant k in Fig3-5, find the work done by the spring when the block (i#) moves from x, to x, Solution: Set origin of the coordinate axis ox at its relaxed position O. dw= F(x)dx==kdx W=F()dx=-kcdx (3-10)F=-kx 2 2红 2 弹性力的功取决于始末位置,与路径无关。 以后知道这种力叫保守力,有相关的势能
Example 3.3 Work done by a spring （弹性力的功） with a force constant k in Fig.3-5, find the work done by the spring when the block（物块） moves from x1 to x2 . Solution: Set origin of the coordinate axis ox at its relaxed position O: F = −kx dW = F(x)dx = −kxdx 2 2 2 1 2 1 2 1 2 1 2 1 kx kx W F x dx kx dx x x x x = − = = − ( ) （3-10） 弹性力的功取决于始末位置，与路径无关。 以后知道这种力叫保守力，有相关的势能