Mechanics Exercise Class Ⅰ

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1 Mechanics Exercise Class Ⅰ

2 英文数字、算式表达法 = 　is equal to / equals >　is larger than < is less than >> is much greater than a = b + c 　　a is equal to b plus c a = b - c 　　 a is equal to b minus c a = b x c 　　a is equal to b times c / a equals to b multiplied by c a = b/c 　　 a is equal to b divided by c / a equals to b over c    (square) root x / the square root of x      cube root (of) x ;  fourth root (of) x ;   nth root (of) x  

3 英文数字、算式表达法 f( x ) fx / f of x / the function f of x
    the limit as x approaches zero      the integral from zero to infinity   x2 　　 x square / x to the second power / x to the power two x3 　　 x cube / x to the third power / x to the power three x the seventh power of x（x to the seventh power） lognX 　log x to the base n

4 Important Formulas Newton’s Second Law Drag Force
Centripetal Acceleration Work and Kinetic Energy Work Done by a Constant Force Work Done by a Spring Force Work Done by a Variable Force

5 52P A model rocket fired vertically from the ground ascends with a constant
vertical acceleration of the 4.00m/s2for 6.00 s. Its fuel is then exhausted ,so it continues upward as a free-fall particle and then falls back down. (a) What is the maximum altitude reached? (b) What is the total time elapsed from the takeoff until the rocket strikes the ground? P30 Solution: (a) One key idea is since the fuel is exhausted and before the rocket strikes the ground, its acceleration is g of magnitude. And when the fuel is exhausted ,the velocity is (upward) y then we can get the position of the rocket Second key idea is the velocity equals zero, when the rocket is at maximum altitude. Using Eq 2-16, we obtain So the maximum altitude is

6 (b) Since the fuel is exhausted and before the rocket returns, the time interval is
Then the rocket from the maximum altitude falls back to the ground. From the Eq. 2-15, we can obtain Work out this equation, yielding So the total time elapsed from the takeoff until the rocket strikes the ground is

7 43p. A block of mass m1 on a frictionless inclined plane of angle is connected
by a cord over a massless , frictionless pulley to a second block of mass m2 hanging vertically. What (a) the magnitude of the acceleration of each block and (b) The direction of the acceleration of the hanging block? (c) what is the tension of the cord? P96 Solution: Choose m2 to be a system and draw it’s free-body diagram With Newton’s second law applied to the m2 system, we can obtain m1 m2 m2 m2g T (1) Choose m1 to be a system and draw it’s free-body diagram and we can write Parallel direction m1g N T Perpendicular direction The acceleration components a1 and a2, have the same value since the string does not stretch, thus (2)

8 Now we can solve equation (1) and (2) simultaneously for T and a2
Now we can solve equation (1) and (2) simultaneously for T and a2. First solve equation (1) for T (3) Then substitute for T into equation (2): Solving for a2, we find Discussion : If the direction of the acceleration of the hanging block is vertically downward. On the other hand if the direction is vertically upward. Substituting for a2 for Eq. (3), we find that tension in the rope is of magnitude

9 The force shown in the Fig. has magnitude Fp=20N and makes an angle
of 300 to the ground .Calculate the work done by this force when the wagon is dragged 100m along the ground. Solution: We choose the x axis horizontal to the right And the y axis vertical upward. Then Whereas d=100m. Then using Eq (Dot product)

10 24E. A 5.0kg block moves in a straight line on a horizontal frictionless surface
Under the influence of a force that varies with position as shown in Fig. 7-31 How much work is done by the force as the block moves from the origin to x=8.0m? P138 Solution: A block moves in a one-dimension line, so The Eq7-36 can be simplified as Key idea: the work done the system by the force component Fx as the system moves from xi to xf is the area under the curve between xi and xf . The work done is the area under the graph between x=0m to x=8.0m is

11 2.8.3一辆卡车在平直路面上以恒定速率30m/s行驶，在此车上射出一抛体，要求在车前进60m时，抛体仍落回到车上原抛出点，问抛体射出时相对于卡车的初速度的大小和方向，空气阻力不计。P59
解，以卡车为参照系，以起抛点为坐标原点，建立直角坐标 系o-xy,如图所示。以抛出时刻为计时起点。 由已知， 得： 表明：抛射体相对卡车以9.8m/s的速率竖直上抛时， 当卡车前进了60m，抛体落回抛射点。

12 3.4.5质量为 的斜面可在光滑的水平面上滑动，斜面倾为 ，质量为 的运动员与斜面之间亦无摩擦，求运动员相对斜面的加速度及其对斜面的压力。P108
解，隔离物体： 对于 对于 联立求解：

13 3.4.9 跳伞运动员初张伞时的速度为 ，阻力大小与速度平方 成正比： ，人伞总质量为m。求 的函数。P109
3.4.9 跳伞运动员初张伞时的速度为 ，阻力大小与速度平方 成正比： ，人伞总质量为m。求 的函数。P109 提示： 积分时可利用式 解： 令 上式写成 积分 代入

14 质量为m=0.5kg的木块可在水平光滑直杆上滑动。木块与一不可伸长的轻绳相连。绳跨过一固定的光滑小环。绳端作用着大小不变的力T=50N.木块在A点时具有向右的速率。求力T将木块自A拉至B点的速度。P152 T A B 解： 做功为零 由动能定理： 式中 A B o 利用积分公式： 则上式